CBSE Class 12 Physics Optics chapter notes and important questions

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OPTICS

Optics – gist of the chapter and question bank with answers

RAY OPTICS – GIST

1. Reflection by convex and concave mirrors. (a) Mirror formula: \(\dfrac1v+\dfrac1u=\dfrac1f\), where \(u\) is the object distance, \(v\) is the image distance and \(f\) is the focal length. (b) Magnification: \(m=-\dfrac vu=\dfrac{f-v}{f}=\dfrac{f}{f-u}\); \(m\) is negative for real images and positive for virtual images.
2. Refraction. (c) A ray of light bends when it enters from one medium into another having a different optical density. (d) The Sun can be seen before the actual sunrise and after the actual sunset due to atmospheric refraction. (e) An object under water (in any medium) appears raised due to refraction when observed along the normal: \(n=\dfrac{\text{real depth}}{\text{apparent depth}}\), where \(n\) is the refractive index of the medium. The shift in the apparent position of the object is \(X=t\left(1-\dfrac1n\right)\), where \(t\) is the actual depth (or thickness) of the medium. (f) Snell’s law: for a given colour of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant when light travels from a rarer to a denser medium, \(\dfrac{\sin i}{\sin r}=\dfrac{n_2}{n_1}\). (g) The absolute refractive index is the ratio of the velocity of light in vacuum to the velocity of light in the medium, \(n=\dfrac cv\). For air \(n=1\).
3. Total internal reflection. (h) When a ray of light travels from a denser to a rarer medium and the angle of incidence is greater than the critical angle, the ray is reflected back into the denser medium. This phenomenon is called total internal reflection, and \(\sin C=\dfrac{n_R}{n_D}\). (i) Diamond has a high refractive index and hence a low critical angle (\(C=24.4^\circ\)). This promotes multiple total internal reflections, causing its brilliance and lustre. Examples of total internal reflection are the formation of mirage and the working of an optical fibre.
4. Refraction at a convex spherical surface. When light falls on a convex refracting surface, it bends, and the relation between \(u\), \(v\) and \(R\) is \(\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}\).
5. Lens maker’s formula (thin lens formula): \(\dfrac1f=\left(\dfrac{n_2-n_1}{n_1}\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\). For a convex lens \(R_1\) is positive and \(R_2\) is negative; for a concave lens \(R_1\) is negative and \(R_2\) is positive. Whether a lens behaves as converging or diverging depends upon the values of \(n_L\) (lens) and \(n_m\) (medium).
6. Lenses in contact. When two lenses are kept in contact, the equivalent focal length is given by \(\dfrac1F=\dfrac1{f_1}+\dfrac1{f_2}\) and \(P=P_1+P_2\).
7. Lens formula: \(\dfrac1v-\dfrac1u=\dfrac1f\).
8. Prism. When light passes through a glass prism it undergoes refraction. The refractive index of the material of the prism is \(n=\dfrac{\sin\left(\dfrac{A+D_m}{2}\right)}{\sin\left(\dfrac A2\right)}\). As the angle of incidence increases, the angle of deviation decreases, reaches a minimum value and then increases. This minimum value is called the angle of minimum deviation \(D_m\).
9. At minimum deviation \(i=e\) and the refracted ray inside the prism is parallel to the base. For a small-angled prism, \(\delta=(n-1)A\).
10. Dispersion. When white light (polychromatic or composite) is passed through a glass prism, it splits up into its component colours (monochromatic). This phenomenon is called dispersion.
11. Rainbow is formed due to the combined effect of refraction and total internal reflection of sunlight by spherical water droplets of rain.
12. Scattering of light takes place when the size of the particle is very small compared to the wavelength of light. The intensity of scattered light is \(I\propto\dfrac1{\lambda^4}\). The following can be explained by scattering: (i) the sky is blue; (ii) the sky is reddish at sunrise and sunset; (iii) infrared photography is used on foggy days; (iv) the orange colour of the black box; (v) yellow light is used in vehicles on foggy days; (vi) red light is used in signals.

OPTICAL INSTRUMENTS – GIST

1. Compound microscope. Objective: the converging lens nearer to the object. Eyepiece: the converging lens through which the final image is seen. Both are of short focal length; the focal length of the eyepiece is slightly greater than that of the objective.
2. Angular magnification (magnifying power) \(M\) of a compound microscope: \(M=M_o\times M_e=\dfrac{v_o}{-u_o}\left(1+\dfrac D{f_e}\right)\), or \(M=\dfrac{-L}{f_o}\left(1+\dfrac D{f_e}\right)\), when the final image is formed at the least distance of distinct vision \(D\). In normal adjustment, i.e. with the final image at infinity, \(M\approx\dfrac{-L}{f_o}\times\dfrac D{f_e}\), where \(L\) is the distance between the objective and the eyepiece (tube length).
3. Astronomical telescope (image formed at infinity – normal adjustment). The focal length of the objective is much greater than that of the eyepiece. The aperture of the objective is also large, to allow more light to pass through it.
4. Angular magnification (magnifying power) of a telescope in normal adjustment: \(M=\dfrac\beta\alpha=\dfrac{-f_o}{f_e}\). (\(f_o+f_e=L\) is called the length of the telescope in normal adjustment.)
5. Newtonian telescope (reflecting type): magnifying power \(M=\dfrac{f_o}{f_e}\); a concave mirror forms the image and a plane mirror reflects it to the eyepiece.
6. Cassegrain telescope: refer to NCERT. Resolving power of a microscope \(=\dfrac1{\Delta d}=\dfrac{2\mu\sin\theta}{\lambda}\). It depends on (i) the wavelength \(\lambda\), (ii) the refractive index of the medium between the object and the objective, and (iii) the half-angle \(\theta\) of the cone of light from the object that enters the objective. Resolving power of a telescope \(=\dfrac1{d\theta}=\dfrac{a}{1.22\lambda}\). It depends on (i) the wavelength \(\lambda\) and (ii) the diameter \(a\) of the objective.

Figure: resolving power of a microscope – object O, objective and half-angle θ of the cone of light entering the objective.

Figure: Newtonian (reflecting) telescope – light from star, concave mirror, plane mirror, eyepiece, eye.

Figure: astronomical telescope in normal adjustment – f_o + f_e = L, angles α and β, image at infinity.

Figure: compound microscope – objective, eyepiece, intermediate image A′B′ and final virtual image A″B″ at distance D; u_o, v_o, F_o, 2F_o, F_e, L.

QUESTION BANK – RAY OPTICS

ONE MARK QUESTIONS

Write the value of the angle of reflection for a ray of light falling normally on a mirror.

Answer: Zero.

How does the dispersive power of a glass prism change when it is dipped in water?

Answer: It decreases.

Light falls from glass to air. Find the angle of incidence for which the angle of deviation is \(90^\circ\) if the refractive index of glass is \(\sqrt2\).

Answer: \(45^\circ\).

The critical angle is given by \(\sin C=\dfrac1{\sqrt2}\), so \(C=45^\circ\). A deviation of \(90^\circ\) means the ray is turned through a right angle, which for a totally reflected ray gives \(180^\circ-2i=90^\circ\), i.e. \(i=45^\circ\), which is exactly the critical angle.

Name the phenomenon due to which one cannot see through fog.

Answer: Scattering.

What is the ratio of \(\sin i\) and \(\sin r\) in terms of velocities in the given figure (a ray passing from a medium of refractive index \(\mu_1\), in which the velocity is \(v_1\), into a medium of refractive index \(\mu_2\), in which the velocity is \(v_2\))?

Answer: \(\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}\).

A biconcave lens of focal length 15 cm is cut into two equal halves along the dotted line shown in the figure (a line perpendicular to the principal axis through the middle of the lens). What will be the new focal length of each half?

Answer: 30 cm.

For the same angle of incidence the angles of refraction in three media A, B and C are \(15^\circ\), \(25^\circ\) and \(35^\circ\) respectively. In which medium would the velocity of light be minimum?

Answer: Medium A.

\(\mu=\dfrac{\sin i}{\sin r}=\dfrac cv\). The smallest angle of refraction gives the largest refractive index, and hence the smallest velocity.

What is the ‘power’ of a plane glass plate?

Answer: Zero.

How does the focal length of a lens change when red light incident on it is replaced by violet light?

Answer: It decreases.

The lower half of a concave mirror is painted black. What effect will this have on the image of an object placed in front of the mirror?

Answer: The intensity of the image will be reduced (in this case to half) but there is no change in the size of the image.

An air bubble is formed inside water. Does it act as a converging lens or a diverging lens?

Answer: Diverging lens.

A water tank is 4 m deep. A candle flame is kept 6 m above the level of water. The refractive index of water is \(\dfrac43\). Where will the image of the candle be formed?

Answer: 6 m below the water level (as given in the source).

The source gives this answer without any explanation. It is the image formed by reflection at the water surface, which acts like a plane mirror and forms the image as far behind the surface as the object is in front of it, i.e. 6 m below the water level. If the question meant the image seen by a fish looking up through the surface (refraction), the image would appear at \(6\times\dfrac43=8\) m above the water level.

A convex lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens in the medium?

Answer: Diverging.

\(\dfrac1f=\left(\dfrac{n_L}{n_m}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\). Here \(\dfrac{n_L}{n_m}=\dfrac{1.5}{1.65}\lt1\), so the bracket is negative and the lens behaves as a diverging lens.

Why does the bluish colour predominate in a clear sky?

Answer: Sunlight is scattered by the air molecules, which are much smaller than the wavelength of light. The intensity of scattered light varies as \(\dfrac1{\lambda^4}\), so blue light (short wavelength) is scattered much more than red light (long wavelength). The scattered blue light reaches our eyes from all directions, so the sky looks blue.

A converging lens is kept coaxially in contact with a diverging lens, both the lenses being of equal focal lengths. What is the focal length of the combination?

Answer: Infinity.

\(\dfrac1F=\dfrac1f+\dfrac1{(-f)}=0\), so \(F=\infty\).

A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens?

Answer: 20 cm.

The image coincides with the object only if the rays are retraced, i.e. they fall normally on the mirror. For this the rays must be parallel between the lens and the mirror, so the object is at the focus of the lens.

TWO AND THREE MARKS QUESTIONS

Deduce the expression for the refractive index of a prism in terms of the angle of the prism and the angle of minimum deviation.

Answer: Let a ray fall on one face of a prism of angle \(A\) at angle of incidence \(i\), be refracted at angle \(r_1\), fall on the second face at angle \(r_2\) inside the prism and emerge at angle \(e\).

Geometry of the prism gives \(r_1+r_2=A\). The total deviation is \(\delta=(i-r_1)+(e-r_2)=i+e-A\).

At minimum deviation the ray passes symmetrically, so \(i=e\) and \(r_1=r_2=r\). Then \(2r=A\) gives \(r=\dfrac A2\), and \(\delta=D_m=2i-A\) gives \(i=\dfrac{A+D_m}{2}\).

By Snell’s law,

\[n=\dfrac{\sin i}{\sin r}=\dfrac{\sin\left(\dfrac{A+D_m}{2}\right)}{\sin\left(\dfrac A2\right)}\]

Define the refractive index of a transparent medium. A ray of light passes through a triangular prism. Plot a graph showing the variation of the angle of deviation with the angle of incidence.

Answer: The refractive index of a transparent medium is the ratio of the speed of light in vacuum to its speed in the medium, \(\mu=\dfrac cv=\dfrac{\sin i}{\sin r}\).

As the angle of incidence increases, the angle of deviation first decreases, reaches a minimum value \(D_m\) (when \(i=e\)) and then increases again, as shown in the graph.

Draw a labelled diagram of an astronomical telescope in the near-point position. Write the expression for its magnifying power.

Answer: In the near-point position the final image is formed at the least distance of distinct vision \(D\) from the eye. The objective forms a real, inverted image of the distant object at a distance a little less than \(f_e\) from the eyepiece, and the eyepiece acts as a magnifier giving a virtual image at \(D\). The magnifying power is

\[M=-\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)\]

(The ray diagram is a standard textbook figure and is not reproduced here.)

A convex lens of focal length \(f_1\) is kept in contact with a concave lens of focal length \(f_2\). Find the focal length of the combination.

Answer: Let an object at distance \(u\) form an image at \(v_1\) by the first lens: \(\dfrac1{v_1}-\dfrac1u=\dfrac1{f_1}\). This image acts as the object for the second lens, which forms the final image at \(v\): \(\dfrac1v-\dfrac1{v_1}=\dfrac1{f_2}\) (thin lenses in contact, so the distances are measured from the common centre).

Adding, \(\dfrac1v-\dfrac1u=\dfrac1{f_1}+\dfrac1{f_2}=\dfrac1F\). Taking \(f_1\) as positive and \(f_2\) as negative (a concave lens),

\[\dfrac1F=\dfrac1{f_1}+\dfrac1{f_2}\ \Rightarrow\ F=\dfrac{f_1f_2}{f_1+f_2}\]

If the magnitudes are used, \(F=\dfrac{f_1f_2}{f_2-f_1}\), and the combination is converging if \(f_2\gt f_1\).

Write the necessary conditions for total internal reflection. Write the relation between the refractive index and the critical angle.

Answer: Light should travel from an optically denser to an optically rarer medium. The angle of incidence must exceed the critical angle.

\[\mu=\dfrac1{\sin i_c}\]

A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging?

Answer: Power of combination \(=-1\) D. The combination is diverging.

\(P=P_1+P_2=\dfrac1{0.25}-\dfrac1{0.20}=4-5=-1\) D. The negative sign shows that it is a diverging system.

A ray of light incident on an equilateral glass prism (\(\mu_g=\sqrt3\)) moves parallel to the base of the prism inside it. Find the angle of incidence for this ray.

Answer: \(\sin i=\sqrt3\sin r=\sqrt3\sin30^\circ=\dfrac{\sqrt3}{2}\), so \(i=\sin^{-1}\left(\dfrac{\sqrt3}2\right)=60^\circ\).

The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If the focal length of the lens is 12 cm, find the refractive index of the material of the lens.

Answer: \(\mu=1.5\).

\[\dfrac1f=(\mu-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\ \Rightarrow\ \dfrac1{12}=(\mu-1)\left(\dfrac1{10}+\dfrac1{15}\right)=(\mu-1)\dfrac16\ \Rightarrow\ \mu=1.5\]

Which two main considerations are kept in mind while designing the objective of an astronomical telescope? Obtain the expression for the magnifying power of an astronomical telescope in normal adjustment position.

Answer: High resolving power and large aperture (large light-gathering power).

In normal adjustment the final image is at infinity. If a distant object subtends an angle \(\alpha\) at the objective, the objective forms its image at its focus, of height \(h\) with \(\alpha\approx\dfrac h{f_o}\). This image lies at the focus of the eyepiece, and the final image subtends an angle \(\beta\approx\dfrac h{f_e}\) at the eye.

\[M=\dfrac\beta\alpha=\dfrac{f_o}{f_e}\]

(the negative sign is used when the inverted image is taken into account, \(M=-\dfrac{f_o}{f_e}\)).

A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.

Answer: Virtual, erect, 30 cm from the pole on the right side of the mirror.

Lens: \(\dfrac1v=\dfrac1f+\dfrac1u=\dfrac1{20}-\dfrac1{60}=\dfrac1{30}\), so \(v=30\) cm to the right of the lens. This point is 15 cm behind the mirror, so it acts as a virtual object for the mirror.Mirror: \(f=+10\) cm (convex), \(u=+15\) cm (virtual object). \(\dfrac1v=\dfrac1f-\dfrac1u=\dfrac1{10}-\dfrac1{15}=\dfrac1{30}\), so \(v=+30\) cm, i.e. a virtual, erect image 30 cm behind the mirror, on its right side.

Draw a schematic labelled ray diagram of a reflecting type telescope. Write two advantages over refracting telescopes.

Answer: A reflecting telescope uses a large concave mirror as the objective; a small plane (or convex) secondary mirror sends the image to an eyepiece. (The ray diagram is a standard textbook figure and is not reproduced here.)

Advantages: (i) it produces a brighter image, as a mirror has a large aperture and large light-gathering power; (ii) a larger focal length gives a larger magnifying power; (iii) there is no chromatic aberration, and the mirror can be supported from behind.

Use the mirror equation to show that an object placed between \(f\) and \(2f\) of a concave mirror produces a real image beyond \(2f\).

Answer: Let the focal length be \(f=-F\) and the object distance \(u=-x\), where \(F\lt x\lt2F\) (Cartesian sign convention). The mirror equation gives \(\dfrac1v=\dfrac1f-\dfrac1u=-\dfrac1F+\dfrac1x=\dfrac{F-x}{Fx}\).

As \(x\gt F\), \(v\) is negative, i.e. the image is real (in front of the mirror). Its distance is \(|v|=\dfrac{Fx}{x-F}\). Now \(|v|\gt2F\) if \(Fx\gt2F(x-F)\), i.e. \(x\lt2F\), which is true. So the image is real and lies beyond \(2f\).

With the help of a suitable diagram, derive the mirror formula for a concave mirror.

Answer: Let AB be an object on the principal axis in front of a concave mirror of pole P, focus F and centre of curvature C. Its real image is A′B′ (the usual diagram has a ray from B parallel to the axis that passes through F after reflection, and a ray from B to P that is reflected at the same angle).

Using magnitudes (distances in front of the mirror taken as positive for the derivation): \(PA=u\), \(PA'=v\), \(PF=f\). Triangles A′B′F and MPF (where PM=AB) are similar, so \(\dfrac{A'B'}{AB}=\dfrac{A'F}{PF}=\dfrac{v-f}{f}\). Triangles A′B′P and ABP are similar, so \(\dfrac{A'B'}{AB}=\dfrac vu\).

Equating, \(\dfrac vu=\dfrac{v-f}{f}\Rightarrow vf=uv-uf\). Dividing by \(uvf\), \(\dfrac1u=\dfrac1f-\dfrac1v\), so

\[\dfrac1u+\dfrac1v=\dfrac1f\]

With the Cartesian sign convention the same relation holds, \(\dfrac1v+\dfrac1u=\dfrac1f\), with \(u\), \(v\) and \(f\) taken with their signs.

A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive index 1.65, (ii) a medium of refractive index 1.33. (a) Will it behave as a converging or a diverging lens in the two cases? (b) How will its focal length change in the two media?

Answer: The focal length in a medium is given by \(\dfrac1{f_m}=\left(\dfrac{n_L}{n_m}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\).

(a) (i) In a medium of refractive index 1.65, \(\dfrac{1.5}{1.65}-1\lt0\), the power becomes negative and the lens is diverging. (ii) In a medium of refractive index 1.33, \(\dfrac{1.5}{1.33}-1\gt0\), the power remains positive and the lens behaves as a converging lens.(b) (i) The focal length becomes negative (about \(-5.5\) times its value in air). (ii) The focal length increases (about \(3.9\) times its value in air).

Use the mirror equation to show that (a) an object placed between \(f\) and \(2f\) of a concave mirror produces a real image beyond \(2f\); (b) a convex mirror always produces a virtual image independent of the location of the object; (c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

Answer:

(a) Shown in the answer to the earlier question: for \(F\lt x\lt2F\), \(v\) is negative (real image) and \(|v|\gt2F\).(b) For a convex mirror \(f\gt0\) and the object is real, so \(u\lt0\). Then \(\dfrac1v=\dfrac1f-\dfrac1u=\dfrac1f+\dfrac1{|u|}\gt0\), so \(v\) is always positive: the image is always virtual (behind the mirror) and lies between the pole and the focus.(c) For a concave mirror let \(f=-F\), \(u=-x\) with \(x\lt F\). Then \(\dfrac1v=-\dfrac1F+\dfrac1x=\dfrac{F-x}{Fx}\gt0\), so \(v\) is positive (virtual image behind the mirror). The magnification \(m=-\dfrac vu=\dfrac{v}{x}=\dfrac{F}{F-x}\gt1\), so the image is erect and enlarged.

A point object O is kept in a medium of refractive index \(n_1\) in front of a convex spherical surface of radius of curvature \(R\) which separates the second medium of refractive index \(n_2\) from the first one. Draw the ray diagram showing the formation of the image and deduce the relationship between the object distance and the image distance in terms of \(n_1\), \(n_2\) and \(R\). Hence obtain the expression for the lens maker’s formula.

Answer: Let the point object O be on the principal axis at distance \(u\) from the pole P of the surface, with centre of curvature C. A ray OA meets the surface at A at a small height \(h\) and, after refraction, meets the axis at the image I at distance \(v\) (assume \(n_2\gt n_1\) and a real image). Let the angles made by OA, AC and AI with the axis be \(\alpha\), \(\gamma\), \(\beta\) respectively.

The angle of incidence is \(i=\alpha+\gamma\) (exterior angle of \(\triangle\)OAC) and the angle of refraction is \(r=\gamma-\beta\) (exterior angle of \(\triangle\)AIC). For small angles, \(\alpha\approx\dfrac hu\) (taking magnitudes), \(\beta\approx\dfrac hv\), \(\gamma\approx\dfrac hR\). Snell’s law for small angles gives \(n_1i=n_2r\), i.e. \(n_1(\alpha+\gamma)=n_2(\gamma-\beta)\).

Substituting, \(n_1\left(\dfrac hu+\dfrac hR\right)=n_2\left(\dfrac hR-\dfrac hv\right)\), which gives, with the Cartesian sign convention (\(u\) negative),

\[\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}\]

Lens maker’s formula. Take a thin lens of refractive index \(n_2\) placed in a medium of index \(n_1\), with surfaces of radii \(R_1\) and \(R_2\). For the first surface, \(\dfrac{n_2}{v_1}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R_1}\). The image at \(v_1\) is the object for the second surface, so \(\dfrac{n_1}{v}-\dfrac{n_2}{v_1}=\dfrac{n_1-n_2}{R_2}\).

Adding, \(n_1\left(\dfrac1v-\dfrac1u\right)=(n_2-n_1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\). When the object is at infinity, \(v=f\), so

\[\dfrac1f=\left(\dfrac{n_2-n_1}{n_1}\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\]

FIVE MARKS QUESTIONS

Derive the mirror formula. Draw a ray diagram to show the image formation by a concave mirror when the object is kept between the focus and the pole. Using this diagram derive the magnification formula.

Answer:

Mirror formula: as derived above, \(\dfrac1v+\dfrac1u=\dfrac1f\) (Cartesian convention, with signs).Object between F and P: the image formed is virtual, erect and enlarged, behind the mirror. (The ray diagram is the standard one: a ray from the top B of the object parallel to the axis is reflected through F, and a ray from B to the pole P is reflected making the same angle with the axis; the reflected rays diverge and appear to meet behind the mirror at B′.)Magnification: let the object height be \(h\) (upwards, positive) and the image height \(h'\). The ray striking the pole P is reflected so that the angle of incidence equals the angle of reflection; the triangles ABP and A′B′P are similar. With the sign convention, \(\dfrac{h'}{h}=-\dfrac vu\).

\[m=\dfrac{h'}{h}=-\dfrac vu\]

A spherical surface of radius of curvature \(R\) separates a rarer and a denser medium. (a) Draw the path of the incident ray of light, showing the formation of a real image. (b) Derive the relation connecting \(u\), \(v\), \(R\) and the refractive indices of the two media. (c) Explain how the focal length of a convex lens changes with increase in wavelength of the incident light.

Answer:

(a), (b) The ray diagram and the derivation are the same as given above for refraction at a spherical surface, giving \(\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}\).(c) The refractive index of glass decreases as the wavelength increases (\(n_{violet}\gt n_{red}\)). By the lens maker’s formula, \(\dfrac1f\propto(n-1)\), so as the wavelength increases, \(n\) decreases and the focal length \(f\) increases.

Derive the lens formula for a concave lens.

Answer: Consider a thin concave lens (refractive index \(n_2\)) in air (\(n_1=1\)) with an object O at distance \(u\) from the optical centre. The lens forms a virtual, erect, diminished image I at distance \(v\) on the same side as the object. Using refraction at the first surface and then at the second surface (as in the lens maker’s derivation): \(\dfrac{n_2}{v_1}-\dfrac1u=\dfrac{n_2-1}{R_1}\) and \(\dfrac1v-\dfrac{n_2}{v_1}=\dfrac{1-n_2}{R_2}\).

Adding, \(\dfrac1v-\dfrac1u=(n_2-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\). For an object at infinity the image is at the focus (\(v=f\)), so \(\dfrac1f=(n_2-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\) and therefore

\[\dfrac1v-\dfrac1u=\dfrac1f\]

For a concave lens \(R_1\) is negative and \(R_2\) is positive, so \(f\) comes out negative.

Two thin convex lenses of focal lengths \(f_1\) and \(f_2\) are placed coaxially in contact. An object is placed at a point beyond the focus of the first lens. Draw a ray diagram to show image formation by the combination and hence derive the expression for the focal length of the combined system.

Answer: The first lens forms a real image \(I_1\) of the object at distance \(v_1\): \(\dfrac1{v_1}-\dfrac1u=\dfrac1{f_1}\). \(I_1\) behaves as the object for the second lens, which forms the final image \(I\) at \(v\): \(\dfrac1v-\dfrac1{v_1}=\dfrac1{f_2}\).

Adding the two equations, \(\dfrac1v-\dfrac1u=\dfrac1{f_1}+\dfrac1{f_2}\). If the combination is replaced by a single lens of focal length \(F\) that forms the image at the same place, \(\dfrac1v-\dfrac1u=\dfrac1F\). Hence

\[\dfrac1F=\dfrac1{f_1}+\dfrac1{f_2}\]

The power of the combination is the sum of the powers, \(P=P_1+P_2\).

(a) Draw a labelled diagram for the formation of an image by a compound microscope. (b) Derive an expression for its magnifying power. (c) Why must the objective and the eyepiece of a compound microscope have short focal lengths?

Answer:

(a) The object AB is placed just beyond the focus \(F_o\) of the objective. The objective forms a real, inverted, magnified image A′B′ inside the focal length of the eyepiece. The eyepiece acts as a simple magnifier and forms a virtual, inverted, highly magnified final image A″B″ at the least distance of distinct vision \(D\) (or at infinity). (The ray diagram is the standard figure of the source and is not redrawn here.)(b) The total magnification is the product of the magnification of the objective and of the eyepiece, \(M=m_o\times m_e\). For the objective, \(m_o=\dfrac{v_o}{-u_o}\approx-\dfrac L{f_o}\), where \(L\) is the separation of the lenses (tube length), since the image lies nearly at the focus of the eyepiece and \(u_o\approx f_o\). For the eyepiece, when the final image is at \(D\), \(m_e=1+\dfrac D{f_e}\); when it is at infinity, \(m_e=\dfrac D{f_e}\). Therefore

\[M=-\dfrac L{f_o}\left(1+\dfrac D{f_e}\right)\ \text{(image at }D\text{)},\qquad M\approx-\dfrac L{f_o}\cdot\dfrac D{f_e}\ \text{(image at infinity)}\]

(c) The magnifying power is proportional to \(\dfrac1{f_o}\) and \(\dfrac1{f_e}\), so to get a large magnification both focal lengths must be small.

(a) Draw a ray diagram to show image formation by an astronomical telescope. (b) Deduce the expression for the magnifying power. (c) Write two limitations.

Answer:

(a) A distant object sends nearly parallel rays to the objective, which forms a real, inverted image at its focus. This image lies at (or just within) the focus of the eyepiece, which forms the final virtual, inverted image (at infinity in normal adjustment). (The standard ray diagram is not redrawn here.)(b) In normal adjustment: \(\alpha\approx\dfrac h{f_o}\), \(\beta\approx\dfrac h{f_e}\), so \(M=\dfrac\beta\alpha=-\dfrac{f_o}{f_e}\) (magnitude \(\dfrac{f_o}{f_e}\)), and the length of the telescope is \(L=f_o+f_e\).(c) Limitations of a refracting telescope: (i) a large objective lens is very heavy, can only be supported at its edges and sags under its own weight; (ii) the image suffers from chromatic and spherical aberrations (and the final image is inverted).

Draw a graph to show the variation of angle of deviation with angle of incidence for a prism. Deduce the expression for the refractive index of the material of the prism.

Answer: The graph is the \(\delta\)–\(i\) curve: \(\delta\) decreases to a minimum \(D_m\) at \(i=e\) and then increases. The expression is \(n=\dfrac{\sin\left(\dfrac{A+D_m}2\right)}{\sin\left(\dfrac A2\right)}\), derived above.

WAVE OPTICS – GIST

Wavefront. A wavelet is the point of disturbance due to propagation of light. A wavefront is the locus of points (wavelets) having the same phase of oscillation. A line perpendicular to a wavefront is called a ‘ray’. A point source gives a spherical wavefront, a linear source gives a cylindrical wavefront, and a very distant source gives a plane wavefront.

Laws of reflection at a plane surface (on Huygens’ principle). Let AB be the incident wavefront, CD the reflected wavefront and XY the reflecting surface. Then \(\sin i-\sin r=0\), i.e. \(\sin i=\sin r\) or \(i=r\).

Laws of refraction at a plane surface (on Huygens’ principle). Let AB be the incident wavefront, CD the refracted wavefront and XY the refracting surface. Then

\[\dfrac{\sin i}{c}-\dfrac{\sin r}{v}=0\quad\text{or}\quad\dfrac{\sin i}{c}=\dfrac{\sin r}{v}\quad\text{or}\quad\dfrac{\sin i}{\sin r}=\dfrac cv=\mu\]

Interference of waves. The waves from \(S_1\) and \(S_2\) reach the point P with some phase difference and hence a path difference

\[\Delta=S_2P-S_1P,\qquad S_2P^2-S_1P^2=\left[D^2+\left(y+\tfrac d2\right)^2\right]-\left[D^2+\left(y-\tfrac d2\right)^2\right]=2yd\]

\[(S_2P-S_1P)(S_2P+S_1P)=2yd\ \Rightarrow\ \Delta\,(2D)=2yd\ \Rightarrow\ \Delta=\dfrac{yd}{D}\]

Comparison of intensities of maxima and minima.

\[\dfrac{I_{max}}{I_{min}}=\dfrac{(a+b)^2}{(a-b)^2}=\dfrac{(a/b+1)^2}{(a/b-1)^2}\]

Relation between intensity (\(I\)), amplitude (\(a\)) of the wave and width (\(w\)) of the slit: \(I\propto a^2\) and \(a\propto\sqrt w\), so

\[\dfrac{I_1}{I_2}=\dfrac{a_1^2}{a_2^2}=\dfrac{w_1}{w_2}\]

Distribution of intensity. The fringes are equally spaced, of equal intensity, and the fringe width is \(\beta=\dfrac{D\lambda}{d}\).

Conditions for sustained interference

1. The two sources producing interference must be coherent.
2. The two interfering wave trains must have the same plane of polarisation.
3. The two sources must be very close to each other and the pattern must be observed at a larger distance to have sufficient width of the fringe (\(\beta=\dfrac{D\lambda}{d}\)).
4. The sources must be monochromatic; otherwise the fringes of different colours will overlap.
5. The two waves must have the same amplitude for better contrast between bright and dark fringes.

Diffraction of light at a single slit. The central maximum is spread on either side of O, so its width is

\[\beta_0=\dfrac{2D\lambda}{d}\qquad(d=\text{width of the slit, } y_1=\dfrac{D\lambda}{d})\]

Fresnel’s distance. \(y_1=\dfrac{D\lambda}{d}\). At Fresnel’s distance, \(y_1=d\) and \(D=D_F\). So \(\dfrac{D_F\lambda}{d}=d\) or

\[D_F=\dfrac{d^2}{\lambda}\]

Malus’ law. When a beam of plane polarised light is incident on an analyser, the intensity \(I\) of light transmitted from the analyser varies directly as the square of the cosine of the angle \(\theta\) between the planes of transmission of the analyser and the polariser.

\[I\propto\cos^2\theta,\qquad I=k(a\cos\theta)^2=ka^2\cos^2\theta\ \Rightarrow\ I=I_0\cos^2\theta\]

(where \(I_0=ka^2\) is the intensity of light transmitted from the polariser).

Figure: Malus’ law – polariser P with amplitude a, analyser A at angle θ; components a cos θ (transmitted) and a sin θ.

Polarisation by reflection and Brewster’s law. When unpolarised light is reflected from a transparent surface at the polarising angle \(\theta_p\), the reflected light is plane polarised and the reflected and refracted rays are perpendicular to each other:

\[\theta_p+r=90^\circ\ \text{ or }\ r=90^\circ-\theta_p,\qquad {}_a\mu_b=\dfrac{\sin\theta_p}{\sin r}=\dfrac{\sin\theta_p}{\sin(90^\circ-\theta_p)}=\tan\theta_p\]

QUESTIONS – WAVE OPTICS

Huygens’ principle

Draw a diagram to show the refraction of a plane wavefront incident on a convex lens and hence draw the refracted wavefront.

Answer: A plane wavefront reaching a convex lens is delayed more at the thick centre than at the thin edges, because light travels more slowly in glass. The central part therefore lags behind and the emerging wavefront is a spherical surface curved towards the focus, i.e. it converges to the focus of the lens. (The diagram shows parallel plane wavefronts on the left and converging spherical wavefronts on the right, with the rays bending towards the focus.)

What type of wavefront will emerge from (i) a point source and (ii) a distant light source?

Answer: (i) A point source emits a spherical wavefront. (ii) A distant source emits a plane wavefront (a very small part of a very large sphere is almost plane).

Define the term wavefront. Using Huygens’ construction draw a figure showing the propagation of a plane wave reflecting at the interface of the two media. Show that the angle of incidence is equal to the angle of reflection.

Answer: A wavefront is the locus of all points which are in the same phase of oscillation.

Let a plane wavefront AB be incident on a reflecting surface XY at angle \(i\). While the point B of the wavefront travels the distance BC to reach the surface in time \(\tau\), the secondary wavelet from A spreads out as a sphere of radius \(AE=v\tau=BC\). The reflected wavefront is the tangent CE from C to this sphere.

In the right-angled triangles ABC and AEC, \(AC\) is common, \(BC=AE\) (both equal \(v\tau\)) and the angles at B and E are right angles, so the triangles are congruent and \(\angle BAC=\angle ECA\). These angles are equal to the angle of incidence \(i\) and the angle of reflection \(r\) respectively. Hence \(i=r\).

Define the term ‘wavefront’. Draw the wavefront and corresponding rays in the case of (i) a diverging spherical wave and (ii) a plane wave. Using Huygens’ construction of a wavefront, explain the refraction of a plane wavefront at a plane surface and hence deduce Snell’s law.

Answer: A wavefront is the locus of points having the same phase of oscillation. (i) For a diverging spherical wave the wavefronts are concentric spheres and the rays are radial lines directed away from the source. (ii) For a plane wave the wavefronts are parallel planes and the rays are parallel straight lines perpendicular to them.

Snell’s law. Let a plane wavefront AB be incident on a plane surface XY separating a rarer medium (speed \(v_1\)) from a denser medium (speed \(v_2\)), making angle \(i\) with the surface. When the point B of the wavefront reaches the surface at C after a time \(\tau\), \(BC=v_1\tau\). In the same time the secondary wavelet from A travels a distance \(AE=v_2\tau\) in the second medium. The refracted wavefront is the tangent CE drawn from C to this wavelet.

In \(\triangle ABC\), \(\sin i=\dfrac{BC}{AC}=\dfrac{v_1\tau}{AC}\). In \(\triangle AEC\), \(\sin r=\dfrac{AE}{AC}=\dfrac{v_2\tau}{AC}\). Dividing,

\[\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}=\dfrac{n_2}{n_1}\]

which is Snell’s law (a constant for the two given media).

Interference

How does the angular separation of interference fringes change, in Young’s experiment, when the distance between the slits is increased?

Answer: When the separation between the slits \(d\) is increased, the fringe width \(\beta=\dfrac{\lambda D}{d}\) decreases, and so the angular separation \(\dfrac\beta D=\dfrac\lambda d\) also decreases.

How does the angular separation of interference fringes in Young’s double-slit experiment change when the distance of separation between the slits and the screen is doubled?

Answer: No effect (the angular separation remains the same).

The angular separation is \(\dfrac\beta D=\dfrac\lambda d\), which does not depend on \(D\).

In a double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is \(0.1^\circ\). What is the spacing between the two slits?

Answer: The spacing between the slits is \(3.44\times10^{-4}\) m.

Angular width \(\theta=\dfrac\lambda d\), so \(d=\dfrac\lambda\theta=\dfrac{600\times10^{-9}}{0.1\times\dfrac\pi{180}}=\dfrac{6\times10^{-7}}{1.745\times10^{-3}}=3.44\times10^{-4}\) m.

If the path difference produced due to interference of light coming out of two slits for yellow colour of light at a point on the screen is \(\dfrac{3\lambda}{2}\), what will be the colour of the fringe at that point? Give reasons.

Answer: The given path difference satisfies the condition for the minimum of intensity for yellow light, \(\Delta=(2n-1)\dfrac\lambda2\) with \(n=2\). Hence when yellow light is used, a dark fringe will be formed at the given point. If white light is used, all the components of white light except the yellow one would be present at that point, so the fringe there is the colour of white light with yellow missing.

State two conditions to obtain sustained interference of light. In Young’s double-slit experiment, using light of wavelength 400 nm, interference fringes of width ‘X’ are obtained. The wavelength of light is increased to 600 nm and the separation between the slits is halved. In order to maintain the same fringe width, by what distance is the screen to be moved? Find the ratio of the distances of the screen in the above two cases.

Answer: Ratio \(D_1:D_2=3:1\); the screen has to be moved towards the slits by \(\dfrac23D_1\).

Conditions: (i) the two sources must be coherent; (ii) the sources must be monochromatic and close together, with waves of nearly equal amplitude.Numerical: \(X=\dfrac{\lambda_1D_1}{d}=\dfrac{400\,D_1}{d}\). In the second case, \(X=\dfrac{\lambda_2D_2}{d/2}=\dfrac{1200\,D_2}{d}\). Equating, \(400D_1=1200D_2\), so \(D_2=\dfrac{D_1}3\) and \(\dfrac{D_1}{D_2}=3:1\). The distance moved is \(D_1-D_2=\dfrac23D_1\).

Two narrow slits are illuminated by a single monochromatic source. Name the pattern obtained on the screen. One of the slits is now completely covered. What is the name of the pattern now obtained on the screen? Draw the intensity pattern obtained in the two cases. Also write two differences between the patterns obtained in the above two cases.

Answer: With both slits open, an interference pattern is obtained. When one slit is covered, a single-slit diffraction pattern is obtained.

Differences: (i) in the interference pattern all bright fringes are of equal width (\(\beta=\dfrac{\lambda D}{d}\)) and nearly equal intensity, whereas in the diffraction pattern the central maximum is twice as wide as the other maxima and is much brighter; (ii) in the interference pattern the intensity of the bright fringes does not fall off appreciably, whereas in the diffraction pattern the intensity of the secondary maxima decreases rapidly on either side of the central maximum.

In Young’s double-slit experiment, monochromatic light of wavelength \(\lambda\) is used. The intensity of light at a point on the screen where the path difference is \(\lambda\) is \(K\) units. What is the intensity of light at a point where the path difference is \(\lambda/3\)?

Answer: \(K/4\).

Let each slit alone give intensity \(I_0\). The resultant intensity is \(I=4I_0\cos^2\dfrac\phi2\). For a path difference \(\lambda\), \(\phi=2\pi\), so \(K=4I_0\). For a path difference \(\dfrac\lambda3\), \(\phi=\dfrac{2\pi}{3}\), so \(I=4I_0\cos^2\dfrac\pi3=4I_0\times\dfrac14=I_0=\dfrac K4\).

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Answer:

(a) \(x=n\lambda_1\dfrac Dd\). For the third bright fringe, \(n=3\): \(x=3\times650\,\dfrac Dd=1950\,\dfrac Dd\) nm.(b) The bright fringes coincide when \(n_1\lambda_1=n_2\lambda_2\), i.e. \(n_1\times650=n_2\times520\), so \(\dfrac{n_1}{n_2}=\dfrac{4}{5}\). The least values are \(n_1=4\), \(n_2=5\): \(x=5\times520\,\dfrac Dd=2600\,\dfrac Dd\) nm.

In a double-slit experiment the angular width of a fringe is found to be \(0.2^\circ\) on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take the refractive index of water to be \(\dfrac43\).

Answer: \(0.15^\circ\).

In water the wavelength becomes \(\lambda'=\dfrac\lambda\mu\), so the angular width \(\theta=\dfrac\lambda d\) becomes \(\theta_2=\dfrac{\theta_1}{\mu}\), i.e. \(\mu=\dfrac{\theta_1}{\theta_2}\). Hence \(\theta_2=\dfrac34\theta_1=\dfrac34\times0.2=0.15^\circ\).

A narrow monochromatic beam of light of intensity \(I\) is incident on a glass plate. Another identical glass plate is kept close to the first one and parallel to it. Each plate reflects 25% of the incident light and transmits the remaining. Calculate the ratio of the minimum and maximum intensity in the interference pattern formed by the two beams obtained after reflection from each plate.

Answer: \(I_{min}:I_{max}=1:49\).

Let \(I\) be the intensity of the beam incident on the first plate. Each plate reflects 25% and transmits 75% of the light falling on it. Beam 1 (incident) \(=I\). Beam 2 (reflected by the first plate) \(=\dfrac{25}{100}I=\dfrac I4\). Beam 3 (transmitted by the first plate) \(=\dfrac{75}{100}I=\dfrac34I\). Beam 4 (reflected by the second plate) \(=\dfrac14\times\dfrac34I=\dfrac3{16}I\). Beam 5 (transmitted back through the first plate) \(=\dfrac34\times\dfrac3{16}I=\dfrac9{64}I\).

The two beams that interfere after reflection are beams 2 and 5. The amplitude ratio is \(R=\sqrt{\dfrac{I_2}{I_5}}=\sqrt{\dfrac{I/4}{9I/64}}=\dfrac43\). Then

\[\dfrac{I_{min}}{I_{max}}=\left(\dfrac{R-1}{R+1}\right)^2=\left(\dfrac{4/3-1}{4/3+1}\right)^2=\dfrac1{49}\]

In a two-slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance \(D\) from the slits. If the screen is moved \(5\times10^{-2}\) m towards the slits, the change in fringe width is \(3\times10^{-5}\) m. If the distance between the slits is \(10^{-3}\) m, calculate the wavelength of the light used.

Answer: \(\lambda=6\times10^{-7}\) m \(=6000\) Å.

The fringe width in the two cases is \(\beta=\dfrac{D\lambda}d\) and \(\beta'=\dfrac{D'\lambda}d\). So \(\beta-\beta'=\dfrac{(D-D')\lambda}{d}\), or \(\lambda=\dfrac{(\beta-\beta')d}{D-D'}\). With \(D-D'=5\times10^{-2}\) m, \(\beta-\beta'=3\times10^{-5}\) m and \(d=10^{-3}\) m, \(\lambda=\dfrac{3\times10^{-5}\times10^{-3}}{5\times10^{-2}}=6\times10^{-7}\) m \(=6000\) Å.

Two sources of intensity \(I\) and \(4I\) are used in an interference experiment. Find the intensity at points where the waves from the two sources superimpose with a phase difference (i) zero (ii) \(\pi/2\) (iii) \(\pi\).

Answer: The resultant intensity at a point where the phase difference is \(\phi\) is \(I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi\). As \(I_1=I\) and \(I_2=4I\), \(I_R=I+4I+2\sqrt{I\cdot4I}\cos\phi=5I+4I\cos\phi\).

(i) When \(\phi=0\), \(I_R=5I+4I\cos0=9I\). (ii) When \(\phi=\dfrac\pi2\), \(I_R=5I+4I\cos\dfrac\pi2=5I\). (iii) When \(\phi=\pi\), \(I_R=5I+4I\cos\pi=I\).

(a) What are coherent sources of light? Two slits in Young’s double-slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength. Why is no interference pattern observed? (b) Obtain the condition for getting dark and bright fringes in Young’s experiment. Hence write the expression for the fringe width. (c) If \(s\) is the size of the source and \(d\) its distance from the plane of the two slits, what should be the criterion for the interference fringes to be seen?

Answer:

(a) Coherent sources are sources that emit light waves of the same frequency with a constant (zero or fixed) phase difference between them. Two different sodium lamps are independent sources: their atoms emit light in random, rapidly changing phases, so the phase difference between the two sources changes randomly and rapidly with time. The bright and dark positions keep shifting too fast to be seen, so no steady interference pattern is observed.(b) The path difference at a point P at distance \(y\) from the centre is \(\Delta=\dfrac{yd}D\). For a bright fringe \(\Delta=n\lambda\), so \(y_n=\dfrac{n\lambda D}d\) (\(n=0,1,2,\dots\)). For a dark fringe \(\Delta=\left(n+\dfrac12\right)\lambda\), so \(y_n=\left(n+\dfrac12\right)\dfrac{\lambda D}d\). The distance between two consecutive bright (or dark) fringes is the fringe width

\[\beta=\dfrac{\lambda D}{d}\]

(c) Fringes are seen clearly if the source is small enough, so that

\[\dfrac sS\lt\dfrac\lambda a\]

where \(s\) is the size of the source, \(S\) its distance from the plane of the two slits and \(a\) the separation of the slits. (In the notation of the question, with \(d\) the distance of the source from the plane of the slits, the criterion is \(\dfrac sd\lt\dfrac\lambda a\).)

What are coherent sources? Why are coherent sources required to produce interference of light? Give an example of interference of light in everyday life. In Young’s double-slit experiment, the two slits are 0.03 cm apart and the screen is placed at a distance of 1.5 m away from the slits. The distance between the central bright fringe and the fourth bright fringe is 1 cm. Calculate the wavelength of light used.

Answer: Coherent sources are two sources of light that emit waves of the same frequency and with a constant phase difference. They are needed because only then does the phase difference at a point stay constant with time, giving fixed positions of constructive and destructive interference (a stable pattern). Example of interference in everyday life: the colours seen in a thin soap film or an oil film on water.

Numerical: the fourth bright fringe is at \(y=\dfrac{4\lambda D}d\), so \(\lambda=\dfrac{yd}{4D}=\dfrac{0.03\times10^{-2}\times1\times10^{-2}}{4\times1.5}=5\times10^{-7}\) m \(=500\) nm.

What is interference of light? Write two essential conditions for a sustained interference pattern to be produced on the screen. Draw a graph showing the variation of intensity versus the position on the screen in Young’s experiment when (a) both the slits are opened and (b) one of the slits is closed. What is the effect on the interference pattern in Young’s double-slit experiment when (i) the screen is moved closer to the plane of the slits, (ii) the separation between the two slits is increased? Explain your answer in each case.

Answer: Interference of light is the phenomenon in which two coherent light waves superpose and the resultant intensity is redistributed in space, with maxima at some points and minima at others.

Two essential conditions: (i) the two sources must be coherent; (ii) the two sources must be monochromatic, close to each other, and of nearly equal amplitude.

(i) When the screen is moved closer, \(D\) decreases, so the fringe width \(\beta=\dfrac{\lambda D}d\) decreases: the fringes become narrower and closer together. (ii) When the slit separation \(d\) is increased, \(\beta\) decreases (\(\beta\propto\dfrac1d\)), so the fringes become closer together.

Diffraction

Why is a coloured spectrum seen when we look through a muslin cloth and not in other clothes?

Answer: Muslin cloth is made of very fine threads and as such fine slits are formed. White light passing through these slits gets diffracted, giving rise to a coloured spectrum. The central maximum is white while the secondary maxima are coloured, because the positions of the secondary maxima (except the central maximum) depend on the wavelength of light. In a coarse cloth the slits formed between the threads are wider and the diffraction is not so pronounced. Hence no such spectrum is seen.

A parallel beam of light of wavelength 600 nm is incident normally on a slit of width \(a\). If the distance between the slit and the screen is 0.8 m and the distance of the 2nd-order maximum from the centre of the screen is 15 mm, calculate the width of the slit.

Answer: \(a=6.4\times10^{-4}\) m \(=0.64\) mm.

Here \(\lambda=600\) nm \(=6\times10^{-7}\) m, \(D=0.8\) m, \(x=15\) mm \(=1.5\times10^{-3}\) m, \(n=2\). Using \(\dfrac xD=\dfrac{n\lambda}a\) (the source treats the 2nd order as \(n=2\)), \(a=\dfrac{n\lambda D}{x}=\dfrac{2\times6\times10^{-7}\times0.8}{1.5\times10^{-3}}=\dfrac{9.6\times10^{-7}}{1.5\times10^{-3}}=6.4\times10^{-4}\) m.

Related note given in the source: interference is due to the superposition of two distinct waves coming from two coherent sources, whereas diffraction is due to the superposition of secondary wavelets generated from different parts of the same wavefront.

Answer the following questions: (a) How do the size and intensity of the central maximum change when the width of the slit is doubled in a single-slit diffraction experiment? (b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? (c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why. (d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily?

Answer:

(a) In a single-slit diffraction experiment, if the width of the slit is made double the original width, the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.(b) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted waves from each slit.(c) Light waves are diffracted from the edge of the circular obstacle, and these waves interfere constructively at the centre of the shadow. This constructive interference produces a bright spot.(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves. The wavelength of light is much smaller than the size of the obstacle, so the diffraction angle is very small and the students cannot see each other. The size of the wall is comparable to the wavelength of sound waves, so the bending of sound takes place at a large angle and the students can hear each other.

Why do light waves not diffract around buildings, while radio waves diffract easily?

Answer: For diffraction to take place the wavelength should be of the order of the size of the obstacle. Radio waves (particularly short radio waves) have wavelengths of the order of the size of buildings and other obstacles in their way, and hence they easily get diffracted. Since the wavelength of light waves is very small, they are not diffracted by buildings.

Draw the diagram showing the intensity distribution of light on the screen for diffraction of light at a single slit. How is the width of the central maximum affected on increasing (i) the wavelength of light used, (ii) the width of the slit? What happens to the width of the central maximum if the whole apparatus is immersed in water, and why?

Answer: The width of the central maximum is \(\beta_0=\dfrac{2\lambda D}{a}\). (i) On increasing the wavelength \(\lambda\), the width of the central maximum increases. (ii) On increasing the slit width \(a\), the width of the central maximum decreases.

When the apparatus is immersed in water, the wavelength becomes \(\lambda'=\dfrac\lambda\mu\) (\(\mu=\dfrac43\)), which is smaller, so the width of the central maximum decreases (becomes \(\dfrac34\) of its value in air).

State the condition under which the phenomenon of diffraction of light takes place. Derive an expression for the width of the central maximum due to diffraction of light at a single slit. A slit of width \(a\) is illuminated by monochromatic light of wavelength 700 nm at normal incidence. Calculate the value of \(a\) for the position of (i) the first minimum at an angle of diffraction of \(30^\circ\); (ii) the first maximum at an angle of diffraction of \(30^\circ\).

Answer:

Condition: diffraction takes place when the size of the obstacle or aperture is comparable to the wavelength of light.Width of the central maximum: let a parallel beam fall on a slit of width \(a\). The first minimum occurs at the angle \(\theta\) for which the path difference between the extreme rays is \(\lambda\), i.e. \(a\sin\theta=\lambda\). For small \(\theta\), \(\theta\approx\dfrac\lambda a\). On a screen at distance \(D\), the first minimum is at \(y_1=D\theta=\dfrac{D\lambda}a\) on either side of the centre O, so the width of the central maximum is

\[\beta_0=2y_1=\dfrac{2D\lambda}{a}\]

(i) For the first minimum, \(a\sin\theta=\lambda\): \(a=\dfrac\lambda{\sin\theta}=\dfrac{700}{\sin30^\circ}=1400\) nm.(ii) For the first maximum, \(a\sin\theta=\dfrac{3\lambda}2\): \(a=\dfrac{3\lambda}{2\sin\theta}=\dfrac{3\times700}{2\times\sin30^\circ}=2100\) nm.

Polarisation

At what angle of incidence should a light beam strike a glass slab of refractive index \(\sqrt3\), such that the reflected and the refracted rays are perpendicular to each other?

Answer: \(i=60^\circ\).

Brewster’s law: \(\tan i_p=\mu=\sqrt3\), so \(i_p=60^\circ\).

What does the statement ‘natural light emitted from the sun is unpolarised’ mean in terms of the direction of the electric vector? Explain briefly how plane polarised light can be produced by reflection at the interface separating two media.

Answer: The statement means that the natural light coming from the sun is a mixture of waves, each having its electric vector directed in a random direction. When light falls on the interface separating two media, the electrons of the second medium start oscillating, which produces the reflected ray in addition to the refracted ray. As light is a transverse wave, only oscillations in the transverse direction produce a light wave; parallel oscillations do not contribute. When a light ray strikes an interface, the component of the electric vector which is parallel to the interface gets reflected. Therefore the reflected light wave is plane polarised.

What is unpolarised light? Explain with the help of a suitable ray diagram how unpolarised light can be polarised by reflection from a transparent medium. Write the expression for the Brewster angle in terms of the refractive index of the denser medium.

Answer: Unpolarised light is light in which the electric vector vibrates in all possible directions in the plane perpendicular to the direction of propagation (like ordinary sunlight).

When unpolarised light is incident on a transparent surface at the polarising (Brewster) angle \(\theta_p\), the reflected light is completely plane polarised, with its electric vector perpendicular to the plane of incidence (parallel to the surface), and the reflected and refracted rays are at \(90^\circ\) to each other. The Brewster angle is given by

\[\tan\theta_p=\mu\]

The critical angle between a given transparent medium and air is denoted by \(i_c\). A ray of light in air enters this transparent medium at an angle of incidence equal to the polarising angle (\(i_p\)). Deduce a relation for the angle of refraction (\(r_p\)) in terms of \(i_c\).

Answer: \(r_p=\tan^{-1}(\sin i_c)\).

For the critical angle, \(\mu=\dfrac1{\sin i_c}\). At the polarising angle, \(\tan i_p=\mu\) and \(i_p+r_p=90^\circ\), so \(\tan r_p=\cot i_p=\dfrac1\mu=\sin i_c\). Hence \(r_p=\tan^{-1}(\sin i_c)\).

What is meant by ‘polarisation’ of a wave? How does this phenomenon help us to decide whether a given wave is transverse or longitudinal in nature?

Answer: Polarisation is the restriction of the vibrations of a wave to a single direction in the plane perpendicular to its direction of propagation. Only transverse waves can be polarised, because in a transverse wave the vibrations are perpendicular to the direction of travel and can be confined to one plane (by passing the wave through a polaroid or a slit). In a longitudinal wave the vibrations are along the direction of travel, so there is no direction perpendicular to it that can be selected. If, when a wave is passed through a polaroid and the polaroid is rotated, the transmitted intensity changes, the wave is transverse; if the intensity stays the same, the wave is longitudinal. This is how polarisation shows that light is a transverse wave and sound in air is a longitudinal wave.

SHORT ANSWER QUESTIONS (3 MARKS)

A converging beam of light is intercepted by a slab of thickness \(t\) and refractive index \(\mu\). By what distance will the convergence point be shifted? Illustrate the answer.

Answer: The convergence point is shifted away from the slab by \(t\left(1-\dfrac1\mu\right)\).

A converging beam would meet at a point P if the slab were not there. The slab makes the rays appear to come from a shorter distance: the thickness \(t\) of glass behaves like an apparent thickness \(\dfrac t\mu\), so the rays are displaced by \(t-\dfrac t\mu=t\left(1-\dfrac1\mu\right)\) in the direction of the beam and meet at a point P′ beyond P. (The illustration is a ray diagram of the converging beam with the slab, the original convergence point P and the shifted point P′.)

In a double-slit experiment \(SS_2\) is greater than \(SS_1\) by \(0.25\lambda\). Calculate the path difference between the two interfering beams from \(S_1\) and \(S_2\) for a maximum at the point P as shown in the figure.

ii) Light rays from inside the glass bend away from the normal and appear to diverge

iii) affection, patience, knowledge about refraction,

Answer: The path difference between the beams from \(S_1\) and \(S_2\) at P is \(S_2P-S_1P=\left(n-\dfrac14\right)\lambda\); for the first maximum (\(n=1\)) it is \(\dfrac{3\lambda}{4}\) (for the central maximum it is \(-\dfrac\lambda4\), i.e. \(\dfrac\lambda4\) in magnitude).

The beam reaching P through \(S_2\) travels the extra distance \(SS_2-SS_1=0.25\lambda\) before the slits. For a maximum at P the total path difference \((SS_2+S_2P)-(SS_1+S_1P)=n\lambda\), so \(S_2P-S_1P=n\lambda-0.25\lambda\).

_{n=REAL DEPTH/APPRENT DEPTH}

and Where in reparative index of medium shift in the position (apparent) of object is X = t {1 – 1/n) where t is the actual depth of the medium

(iii) Infra-red photography used in foggy days.

(iv) Orange colour of black Box