CBSE Class 12 Physics Optical Instrument Theory and Examples

CBSE Class 12 Physics Optical Instrument Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Optical Instruments

1. INTRODUCTION

1.1 Definition

Optical instruments are used primarily to assist the eye in viewing an object.

1.2 Types of Instruments

Depending upon the use, optical instruments can be categorised in the following way:

2. FILM OR SLIDE PROJECTOR

It projects a real, inverted and magnified image of an object, when the object is placed between F & 2F and the screen between 2F and ∞.

2.1 Magnification

If the sides of the slide are \(a\) and \(b\) and the magnification of the projection lens is \(m\), each side will become \(m\) times, so that the area of the image formed is

\[A_i=ma\times mb=m^2ab=m^2A_0\]

2.2 Intensity

As the area of the image becomes \(m^2\) times that of the object, the intensity of the image will become \(1/m^2\) times that of the object, as

\[I A_0=I_i A_i=I_i m^2 A_0\ \Rightarrow\ I_i=\frac{I}{m^2}\]

Examples based on Projector

Example: A slide projector lens has a focal length 10 cm. It throws an image of a 2 cm × 2 cm slide on a screen 5 m from the lens. Find (a) the size of the picture on the screen and (b) the ratio of illumination of the slide and the picture on the screen.

Solution: (a) Here \(f=10\text{ cm}\) and \(v=5\text{ m}=500\text{ cm}\). From the lens formula \(\frac1v-\frac1u=\frac1f\), we have

\[\frac1{500}-\frac1u=\frac1{10}\quad\text{i.e.,}\quad u=-\left[\frac{500}{49}\right]\text{cm}\]

So that

\[m=\frac vu=\frac{500}{(-500/49)}=-49\]

Here the negative sign means the image is inverted with respect to the object. Now as here the object is (2 cm × 2 cm), the size of the picture on the screen is

\[A_i=(2\times49\text{ cm})\times(2\times49\text{ cm})=98\times98\text{ cm}^2\]

(b) As the light energy passing per second through the slide is equal to that in the picture on the screen,

\[IA=I_iA_i\ \Rightarrow\ \frac{I}{I_i}=\frac{A_i}{A}=\frac{(ma)\times(mb)}{(a\times b)}=m^2\]

\[\text{i.e.,}\quad\frac{I}{I_i}=(-49)^2=(49\times49)\]

i.e., the intensity in the picture on the screen will be \(\left[\dfrac1{49\times49}\right]\) times less than that on the slide. This is why, in the case of a projector for observing the image on a screen, the source of light must be very powerful and the room dark.

3. CAMERA

A camera is of two types, viz. pinhole camera and lens camera.

3.1 Pinhole Camera

It is based on the rectilinear propagation of light and forms a real, inverted image on the screen.

\[\theta=\frac{h_0}{u}=\frac{h_i}{v}\ \Rightarrow\ \frac{h_i}{h_0}=\frac vu\]

Note:

The image formed on the screen is neither a shadow (as it is not dark) nor a true image (as the rays do not intersect each other and cannot be seen as an aerial image in the absence of the screen).

3.2 Lens Camera

In it a converging lens whose aperture and distance from the film can be adjusted is used. Usually the object is real and between ∞ and 2F; so the image is real, inverted, diminished and between F and 2F.

(A) Exposure:

For a particular film, a definite amount of light energy must be incident on the film for proper exposure, i.e.,

\[I\times S\times t=\text{constant}\]

where \(I\) = intensity of light, \(S\) = area of the aperture of the lens and \(t\) is the exposure time. i.e. \(I D^2 t=\text{constant}\), where \(D\) = aperture.

(B) f-number:

It is the ratio of the focal length to the aperture of the lens, i.e. \(f\text{-number}=f/D\).

Examples based on Camera

Example: A pinhole camera with a 75 mm high film is to be used to take a picture of a 10 m high tree. The film is 150 mm from the pinhole. How far should the camera be from the tree to include the full height of the tree?

Solution: In the case of a pinhole camera

\[\theta=\frac Iv=\frac Ou\quad\text{i.e.,}\quad u=v\times\frac OI\]

Here \(I=75\text{ mm}\), \(O=10\text{ m}\) and \(v=150\text{ mm}\). So

\[u=(10\text{ m})\times\frac{150\text{ mm}}{75\text{ mm}}=20\text{ m}\]

4. MICROSCOPE

It is an optical instrument used to increase the visual angle of near objects which are too small to be seen by the naked eye.

Microscopes are of two types, viz. simple microscope and compound microscope.

4.1 Simple Microscope

It is also known as a magnifying glass or magnifier and consists of a convergent lens with the object between its focus and optical centre and the eye close to it. The image formed by it is erect, virtual, enlarged and on the same side of the lens as the object, between the object and infinity.

Here

\[\text{Magnifying power}=\frac{\text{visual angle with instrument}}{\text{Maximum visual angle for unaided eye}}=\frac{\theta}{\theta_0}\]

Now \(\theta=\dfrac{h_i}{v}=\dfrac{h_0}{u}\), with \(\theta_0=h_0/D\); so

\[M.P.=\frac{\theta}{\theta_0}=\frac{h_0}{u}\times\frac{D}{h_0}=\frac Du\]

Now, two possibilities are there:

(A) Image is at infinity (far point)

If \(v=\infty\), \(u=f\) (from the lens formula). So

\[M.P.=\frac Du=\frac Df\]

Note:

Here a parallel beam of light enters the eye, i.e., the eye is least strained.

(B) Image is at D (near point)

In this situation \(v=-D\), so that

\[\frac1{-D}-\frac1{-u}=\frac1f\ \ \text{(magnitudes of distances)}\]

or

\[\frac Du=1+\frac Df\]

So

\[M.P.=1+\frac Df\]

Note:

Here the final image is closest to the eye, i.e., the eye is under maximum strain.

Examples based on Microscope

Example: A man with normal near point (25 cm) reads a book with small print using a magnifying glass, a thin convex lens of focal length 5 cm. (a) What are the closest and farthest distances at which he can read the book when viewing through the magnifying glass? What are the maximum and minimum M.P. possible using the above simple microscope?

Solution: (a) As for a normal eye the far and near points are \(\infty\) and 25 cm respectively, so for the magnifier \(v_{max}=-\infty\) and \(v_{min}=-25\text{ cm}\). However, for a lens

\[\frac1v-\frac1u=\frac1f\quad\text{i.e.,}\quad u=\frac{f}{(f/v)-1}\]

So \(u\) will be minimum when \(v=\min=-25\text{ cm}\), i.e.,

\[(u)_{min}=\frac{5}{-(5/25)-1}=-\frac{25}{6}=-4.17\text{ cm}\]

And \(u\) will be maximum when \(v=\max=\infty\), i.e.,

\[(u)_{max}=\frac{5}{(5/\infty)-1}=-5\text{ cm}\]

So the closest and farthest distances of the book from the magnifier (or eye) for clear viewing are 4.17 cm and 5 cm respectively.

(b) In the case of a simple magnifier \(M.P.=D/u\). So M.P. will be minimum when \(u\) = max = 5 cm, i.e.,

\[(M.P.)_{min}=\frac{-25}{-5}=5\times\ \left[=\frac Df\right]\]

And M.P. will be maximum when \(u\) = min = (25/6) cm, i.e.,

\[(M.P.)_{max}=\frac{-25}{-(25/6)}=6\times\ \left[=1+\frac Df\right]\]

Example: If the focal length of a magnifier is 5 cm, calculate (a) the power of the lens, (b) the magnifying power of the lens for relaxed and strained eye.

Solution: (a) The power of a lens is the reciprocal of its focal length in metres,

\[P=\frac{1}{(5\times10^{-2}\text{ m})}=\frac1{0.05}\text{ dioptre}=20\text{ D}\]

(b) For a relaxed eye, M.P. is minimum and will be

\[M.P.=\frac Df=\frac{25}{5}=5\times\]

While for a strained eye, M.P. is maximum and will be

\[M.P.=1+\frac Df=1+5=6\times\]

4.2 Compound Microscope

It consists of two convergent lenses of short focal lengths and apertures arranged co-axially. Lens \(f_0\) is the objective or field lens and \(f_e\) is the eye-piece or ocular. The objective has a smaller aperture and focal length than the eye-piece. The separation between objective and eye-piece can be varied.

\[\text{Magnifying Power}=\frac{\theta}{\theta_0}=\frac{h_i}{u_e}\times\frac{D}{h_0}=\frac{h_i}{h_0}\times\frac D{u_e}\]

But for the objective, \(m=\dfrac vu\), i.e., \(\dfrac{h_i}{h_0}=-\dfrac vu\), so

\[M.P.=-\frac vu\left[\frac D{u_e}\right]\]

where \(u_e\) is the distance of the intermediate image from the eye-piece, and the length of the microscope is \(L=v+u_e\).

Now two possibilities are there:

(A) Final image is at infinity (far point)

\[u_e=f_e\ \Rightarrow\ M.P.=-\frac vu\left[\frac D{f_e}\right]\qquad\text{where }L=v+f_e\]

Note:

A microscope is usually considered to operate in this mode unless stated otherwise.

(B) Final image is at D (near point)

For the eye-piece, \(v_e=D\) (in magnitude), so

\[\frac1{-D}-\frac1{-u_e}=\frac1{f_e}\ \Rightarrow\ \frac1{u_e}=\frac1D\left(1+\frac D{f_e}\right)\]

(magnitudes of \(D\) and \(u_e\) are used in this step, as in the simple microscope)

\[\therefore\ M.P.=-\frac vu\left(1+\frac D{f_e}\right)\quad\text{with}\quad L=v+\frac{f_eD}{f_e+D}\]

Note:

In case \(u\approx f_0\) and \(L=v+u_e\approx v\), so that

\[|M.P.|\approx\frac{L}{f_0}\times\frac D{f_e}\]

IMPORTANT POINTS

1. As the magnifying power is negative, the image seen in a microscope is always truly inverted, i.e., left is turned right with upside down simultaneously.
2. Resolving Power: The minimum distance between two lines at which they are just distinct is called the limit of resolution and the reciprocal of the limit of resolution is called resolving power.\[\therefore\ R.P.=\frac1{\Delta x}=\frac{2\mu\sin\theta}{\lambda}\ \propto\ \frac1\lambda\]

5. TELESCOPE

It is an optical instrument used to increase the visual angle of distant large objects.

Telescopes are mainly of two types, viz. astronomical and terrestrial.

5.1 Astronomical Telescope

It consists of two converging lenses placed coaxially, with the objective having a large aperture and a large focal length, while the eye-piece has a smaller aperture and focal length. The separation between eye-piece and objective can be varied.

\[\text{Magnifying power}=\frac{\text{visual angle with instrument}}{\text{visual angle for unaided eye}}=\frac\theta{\theta_0}\]

\(\theta_0=h/f_0\) & \(\theta=h/(-u_e)\)

\[\Rightarrow\ M.P.=-\left[\frac{f_0}{u_e}\right]\qquad\text{with}\quad L=f_0+u_e\]

Now two possibilities are there:

(A) Final image at infinity (far point)

Here \(v=\infty\) ⇒ \(u_e=f_e\). So

\[M.P.=-(f_0/f_e)\qquad\text{with}\quad L=f_0+f_e\]

Note:

Usually, a telescope operates in this mode unless stated otherwise.

(B) Final image is at D (near point)

Here \(v=D\) ⇒

\[\frac1{-D}-\frac1{-u_e}=\frac1{f_e}\ \Rightarrow\ \frac1{u_e}=\frac1{f_e}\left[1+\frac{f_e}{D}\right]\]

(magnitudes used in this step.) So

\[M.P.=-\frac{f_0}{f_e}\left[1+\frac{f_e}{D}\right]\qquad\text{with}\quad L=f_0+\frac{f_eD}{f_e+D}\]

Note:

1. The above discussion is that of the refracting telescope.
2. Reflecting Telescope: If the field lens of a refracting telescope is replaced by a converging mirror, then the telescope becomes a reflecting one, where \(M.P.=f_0/f_e\).
3. Infrared Telescope: It is used to see distant objects in the dark.

5.2 Terrestrial Telescope

If a lens of short focal length \(f\) is placed at \(2f\) from the intermediate image and this image is at a distance \(2f\) on the other side of it, this image will act as an object for the eye-lens, which will produce an erect image with respect to the object. This lens is called the erecting lens and, as for \(m=-1\), the M.P. and length of the telescope for a relaxed eye will be

\[M.P.=-\frac{f_0}{f_e}(-1)=\frac{f_0}{f_e},\qquad L=f_0+f_e+4f\]

(A) Galilean Telescope

Here the convergent eye-piece of the astronomical telescope is replaced by a divergent lens. Here \(M.P.=f_0/f_e\) with \(L=f_0-f_e\).

Note:

In this telescope, as the intermediate image is outside the tube, the telescope cannot be used for measurements. This was not the case for all previous telescopes.

(B) Binocular

When two telescopes are mounted parallel to each other so that an object can be seen by both the eyes simultaneously, the arrangement is called a binocular. Here, the length of each tube is reduced by using a set of totally reflecting prisms which provide intense, erect image free from lateral inversion. A binocular gives proper 3-D image.

Examples based on Telescope

Example: An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and eye-piece is 36 cm and the final image is formed at infinity. Determine the focal length of the objective and eye-piece.

Solution: In the case of an astronomical telescope, if the object and the final image are both at infinity, \(M.P.=-(f_0/f_e)\) and \(L=f_0+f_e\). So here \(-(f_0/f_e)=-5\) and \(f_0+f_e=36\). Solving these for \(f_0\) and \(f_e\), we get

\[f_0=30\text{ cm}\quad\text{and}\quad f_e=6\text{ cm}\]

Example: A telescope has an objective of focal length 50 cm and an eye-piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distinct vision on a scale 2 m away from the objective. Calculate (a) the magnification produced and (b) the separation between objective and eye-piece.

Solution: As the objective has focal length 50 cm and the object is 2 m from it, it will form the image of the object at a distance \(v\) such that

\[\frac1v-\frac1{-200}=\frac1{50}\quad\text{i.e.,}\quad v=\frac{200}{3}\text{ cm}\]

with

\[m_0=\frac vu=\frac{(200/3)}{-200}=-\frac13\]

and as the focal length of the eye-piece is 5 cm and it forms the image 25 cm in front of it, the distance of the object (image formed by the objective) from it will be

\[\frac1{-25}-\frac1{u_e}=\frac15\quad\text{i.e.,}\quad u_e=-\frac{25}{6}\text{ cm}\]

with

\[m_e=\frac{v_e}{u_e}=-\frac{25}{-(25/6)}=6\]

and hence

(a) Magnification \(m=m_0\times m_e=\left(-\dfrac13\right)\times6=-2\), i.e., the final image is inverted, virtual, double the size of the object and is at a distance of 25 cm in front of the eye lens.

(b) As the distance of the intermediate image (which is between the two lenses) from the objective is \(\dfrac{200}{3}\) cm and from the eye-lens is \(\dfrac{25}{6}\) cm, the separation between objective and eye-piece is

\[L=\frac{200}{3}+\frac{25}{6}=\frac{425}{6}=70.83\text{ cm}\]