CBSE Class 12 Physics Optics Basic Problems

CBSE Class 12 Physics Optics Basic Problems. 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.

1. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that the end closer to the pole is 20 cm away from it. Find the length of the image. [3 Marks]

Answer:
1. Focal length of the concave mirror, f = -10 cm.
2. For the closer end of the rod, object distance u₁ = -20 cm. Using the mirror formula \(\frac{1}{f} = \frac{1}{v_1} + \frac{1}{u_1}\), we get \(\frac{1}{-10} = \frac{1}{v_1} + \frac{1}{-20}\), which gives v₁ = -20 cm.
3. For the farther end of the rod, object distance u₂ = -(20 + 10) = -30 cm. Using the mirror formula, \(\frac{1}{-10} = \frac{1}{v_2} + \frac{1}{-30}\), which gives v₂ = -15 cm.
4. The length of the image is the distance between the two image positions, \(\text{Length} = |v_1 - v_2| = |-20 - (-15)| = 5\) cm.

Teacher's Note:
a) Apply the mirror formula separately for both endpoints since magnification formula applies only for infinitesimally small objects.
b) A common mistake is using the average object distance with longitudinal magnification \(m^2\), which fails here as the rod is not very small.
c) The negative signs for distances must be strictly maintained according to the sign convention.

2. At what distance from a convex mirror of focal length 2.5 m should a boy stand so that his image has a height equal to half the original height? The principal axis is perpendicular to the height. [2 Marks]

Answer:
1. Focal length of the convex mirror, f = +2.5 m, and magnification m = +\(\frac{1}{2}\).
2. Magnification for a mirror is given by m = \(-\frac{v}{u}\), so \(\frac{1}{2} = -\frac{v}{u}\), which gives v = \(-\frac{u}{2}\).
3. Using the mirror formula \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\), substitute v and f to get \(\frac{1}{2.5} = \frac{1}{-u/2} + \frac{1}{u} = -\frac{2}{u} + \frac{1}{u} = -\frac{1}{u}\).
4. Solving for u gives u = -2.5 m, so the boy should stand at a distance of 2.5 m from the convex mirror.

Teacher's Note:
a) Convex mirrors always form virtual, erect, and diminished images behind the mirror.
b) Remember that focal length and magnification signs are crucial for correct results.

3. A 2.0 cm high object is placed on the principal axis of a concave mirror at a distance of 12 cm from the pole. If the image is inverted, real and 5.0 cm high, find the location of the image and the focal length of the mirror [3 Marks]

Answer:
1. Object height \(h_1 = +2.0\) cm, object distance u = -12 cm, and image height \(h_2 = -5.0\) cm (since the image is real and inverted).
2. Magnification m = \(\frac{h_2}{h_1} = \frac{-5.0}{2.0} = -2.5\). Also, m = \(-\frac{v}{u}\), so -2.5 = \(-\frac{v}{-12}\), which gives v = -30 cm.
3. Using the mirror formula \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-30} + \frac{1}{-12} = \frac{-2 - 5}{60} = \frac{-7}{60}\).
4. Focal length f = \(-\frac{60}{7}\) cm = -8.57 cm, and the location of the image is 30 cm in front of the mirror.

Teacher's Note:
a) The magnification formula links object height and distance directly to image parameters.
b) Real images formed by concave mirrors are always inverted, leading to a negative image height.
c) Always retain fractions until the final calculation step for better precision.

4. A point object O is placed in front of a transparent slab at a distance x from its closer surface. It is seen from the other side of the slab by light incident nearly normally to the slab. The thickness of the slab is t and its refractive index is μ. Show that the apparent shift in the position of the object is independent of x and find its value. [3 Marks]

Answer:
1. Apparent shift due to a refractive slab of thickness t and refractive index \(\mu\) is given by the standard formula \(\Delta t = t \left(1 - \frac{1}{\mu}\right)\).
2. The shift depends only on the thickness t of the slab and its refractive index \(\mu\).
3. Since the expression for apparent shift does not contain the object distance x, it is completely independent of x.

Teacher's Note:
a) The shift formula is derived by equating real depth to apparent depth using Snell's law under paraxial approximation.
b) Students often mistakenly assume shifting changes with distance x from the surface.
c) This is a standard theoretical proof expected directly in board examinations.

5. Consider the situation shown in figure. A plane mirror is fixed at a height h above the bottom of a beaker containing water (refractive index \(\mu\)) upto a height d. Find the position of the image of the bottom formed by the mirror. [3 Marks]

[Figure: A beaker containing water up to height d, with a plane mirror fixed horizontally at a height h from the bottom. Rays reflect from the mirror.]

Answer:
1. The apparent depth of the bottom of the beaker as viewed from the water surface is \(\frac{d}{\mu}\).
2. The bottom is at depth d in water, so it appears at depth \(\frac{d}{\mu}\) below the water surface. The water surface is (h - d) below the mirror, so the apparent distance of the bottom from the mirror is \((h - d) + \frac{d}{\mu}\).
3. The plane mirror forms the image at an equal distance behind the mirror. Hence, the image of the bottom is formed at a distance \(h - d + \frac{d}{\mu}\) behind the mirror.

Teacher's Note:
a) Plane mirror image distance equals object distance measured from the mirror surface.
b) Refraction occurs at the water-air boundary when viewed from outside, but here the mirror is inside the water medium.
c) Drawing a clean ray diagram helps clarify the medium boundaries and positions.

6. A concave mirror of radius 40 cm lies on a horizontal table and water is filled in it up to a height of 5.00 cm. A small dust particle floats on the water surface at a point P vertically above the point of contact of the mirror with the table. Locate the image of the dust particle as seen from a point directly above it. The refractive index of water is 1.33. [4 Marks]

[Figure: A concave mirror resting on a table, filled with water up to height 5 cm, with a particle at point P on the surface.]

Answer:
1. Radius of curvature R = -40 cm, so focal length f = R / 2 = -20 cm.
2. The dust particle at P acts as an object for the combination of refraction at the water surface and reflection at the concave mirror.
3. Light travels from water to the mirror, reflects, and refracts back out of water. Using the principle of reversibility or direct ray tracing, the mirror forms an image of P at distance u = -5 cm. Using mirror formula: \(\frac{1}{-20} = \frac{1}{v_m} + \frac{1}{-5}\), which gives \(v_m = +6.67\) cm (virtual image behind the mirror).
4. The virtual image is 6.67 cm behind the mirror, i.e. 5 + 6.67 = 11.67 cm below the water surface. Seen from above, its apparent depth is 11.67/\(\mu\), which is about 8.75 cm (taking \(\mu = 4/3\)). Hence the image is about 8.75 cm below the water surface.

Teacher's Note:
a) This complex problem combines refraction at a plane surface with reflection from a curved mirror.
b) Break the problem into stages: first consider the mirror action, then the refraction of the reflected rays through water.
c) Check all boundary interfaces carefully to assign proper signs.

7. An object is placed 21 cm in front of a concave mirror of radius of curvature 20 cm. A glass slab of thickness 3 cm and refractive index 1.5 is placed close to the mirror in the space between the object and the mirror. Find the position of the final image formed. The distance of the nearer surface of the slab from the mirror is 10 cm. [4 Marks]

Answer:
1. Radius of curvature R = -20 cm, so focal length f = -10 cm.
2. The glass slab introduces an apparent shift given by \(\Delta t = t \left(1 - \frac{1}{\mu}\right) = 3 \left(1 - \frac{1}{1.5}\right) = 3 \left(\frac{1}{3}\right) = 1\) cm towards the mirror.
3. Without the slab, the object distance is u = -21 cm. With the slab shift of 1 cm, the effective object distance for the mirror becomes u' = -(21 - 1) = -20 cm.
4. Using the mirror formula \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u'}\), we have \(\frac{1}{-10} = \frac{1}{v} + \frac{1}{-20}\), which yields v = -20 cm, i.e. the apparent object lies at the centre of curvature of the mirror, so the rays retrace their path through the slab. Hence the final image is formed at the object itself.

Teacher's Note:
a) The presence of a dielectric or refractive slab shifts the effective position of the object.
b) Calculate the slab shift first, then adjust the object distance before applying the mirror formula.
c) Verify that object and image distances are measured from the pole of the mirror.

8. There is a small air bubble inside a glass sphere (\(\mu = 1.5\)) of radius 10 cm. The bubble is 4.0 cm below the surface and is viewed normally from the outside. Find the apparent depth of the bubble. [2 Marks]

Answer:
1. Refractive index of denser medium (glass) \(\mu_1 = 1.5\), refractive index of rarer medium (air) \(\mu_2 = 1.0\).
2. Real depth (distance from the surface) u = -4.0 cm, and radius of curvature R = -10 cm.
3. Using refraction at a spherical surface formula: \(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\).
4. Substituting the values: \(\frac{1.0}{v} - \frac{1.5}{-4.0} = \frac{1.0 - 1.5}{-10}\), which gives apparent depth v = -3.0 cm from the surface.

Teacher's Note:
a) Refraction at spherical surfaces requires careful use of refractive index ratios and sign conventions.
b) The object is inside the sphere, so the center of curvature lies in the opposite direction.

9. A point source S is placed at a distance of 15 cm from a converging lens of focal length 10 cm on its principal axis. Where should a diverging mirror of focal length 12 cm be placed so that a real image is formed on the source itself. [4 Marks]

Answer:
1. For the converging lens, object distance u = -15 cm and focal length f = +10 cm.
2. Using lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), we get \(\frac{1}{10} = \frac{1}{v} - \frac{1}{-15}\), which gives v = +30 cm.
3. For the final image to form on the source itself, the rays must retrace their path, meaning they must fall normally (retrace) on the diverging mirror.
4. This requires the rays to appear to converge at the center of curvature of the diverging mirror. Since the mirror's focal length is f_m = -12 cm, its radius of curvature is R = 24 cm, so the mirror must be placed 30 - 24 = 6 cm to the right of the lens.

Teacher's Note:
a) Reversibility of light is the key concept: rays must strike the mirror normally to retrace their path.
b) Combining lenses and mirrors requires treating the image of the first optical component as the virtual or real object for the second.

10. A converging lens of focal length 15 cm and a converging mirror of focal length 20 cm are placed with their principal axes coinciding. A point source S is placed on the principal axis at a distance of 12 cm from the lens on the other side. It is found that the final beam of light comes out parallel to the principal axis. Find the separation between the mirror and the lens. [4 Marks]

Answer:
1. For the converging lens, f = +15 cm and u = -12 cm.
2. Using the lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), we find \(\frac{1}{15} = \frac{1}{v} - \frac{1}{-12}\), which gives v = -60 cm (virtual image formed at 60 cm in front of the lens).
3. For the final beam to emerge parallel after passing back through the lens, the mirror must form its image at the focus of the lens, 15 cm from the lens. Let the lens-mirror separation be L. The object for the mirror is the virtual image, at distance L + 60 from the mirror, and its image must be at distance L - 15 from the mirror.
4. Mirror formula: \(\frac{1}{L-15} + \frac{1}{L+60} = \frac{1}{20}\), which gives \(L^2 + 5L - 1800 = 0\), so L = 40 cm. The separation between the mirror and the lens is 40 cm.

Teacher's Note:
a) Parallel emerging rays mean the source/object point for the final element lies at its focus.
b) Systematic tracking of intermediate image positions prevents sign convention errors in multi-component optical systems.

11. A biconvex thin lens is prepared from glass (\(\mu = 1.5\)), the two bounding surfaces having equal radii of 25 cm each. One of the surfaces is silvered from outside to make it reflecting. Where should an object be placed before this lens so that the image is formed on the object itself ? [5 Marks]

Answer:
1. The focal length f of the unsilvered biconvex lens is given by \(\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).
2. Substituting the given values, \(\frac{1}{f} = (1.5 - 1)\left(\frac{1}{25} - \frac{1}{-25}\right) = 0.5 \times \frac{2}{25} = \frac{1}{25}\), so \(f = 25 \text{ cm}\).
3. The equivalent focal length \(F\) of the silvered lens system is given by \(\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m}\), where \(f_l = f = 25 \text{ cm}\) and the power of the equivalent mirror is combined.
4. Since the second surface is silvered, it acts as a concave mirror of radius \(R = 25 \text{ cm}\), so its focal length is \(f_m = \frac{R}{2} = 12.5 \text{ cm}\).
5. Thus, \(\frac{1}{F} = \frac{2}{25} + \frac{1}{12.5} = \frac{2}{25} + \frac{2}{25} = \frac{4}{25}\), which gives \(F = \frac{25}{4} = 6.25 \text{ cm}\).
6. For the image to coincide with the object, the rays must retrace their path, meaning the object must be placed at the center of curvature of the equivalent mirror, at a distance of \(2F = 12.5 \text{ cm}\) in front of the lens.

Teacher's Note:
a) Combination of lens and mirror formulas determines the equivalent focal length.
b) Pay close attention to sign conventions for radii of curvature and focal lengths.
c) Reversing light rays happens when rays fall normally on a mirror surface.
d) The resulting system behaves entirely as a concave mirror of focal length \(6.25 \text{ cm}\).

12. Consider the situation shown in figure. The bottom of the vessel is a reflecting plane mirror, S is a small fish and T is the observer's eye. Refractive index of water is \(\mu\).
(a) At what distance(s) from itself will the fish see image(s) of the eye [3 Marks]
(b) At what distances from itself will the eye see the image(s) of the fish. [3 Marks]

[Figure: A vessel filled with water of height H. The bottom is a plane mirror. A fish S is at depth H/2 from the surface, and observer's eye T is at height H above the water surface.]

Answer:
(a) 1. The fish sees the eye directly through air and water, and also through reflection from the bottom mirror.
2. For direct vision, apparent depth of the eye from the water surface is \(\mu H\), so the distance from the fish (at depth \(\frac{H}{2}\)) to the direct image of the eye is \(\frac{H}{2} + \mu H = H\left(\mu + \frac{1}{2}\right)\) above itself.
3. For the reflected image, the eye forms a virtual image below the mirror; its apparent distance below the bottom is \(H + \mu H\), so the fish sees it at distance \(\frac{H}{2} + H + \mu H = H\left(\mu + \frac{3}{2}\right)\) below itself.
(b) 1. The eye sees the fish directly and through reflection from the mirror.
2. For direct vision, the apparent depth of the fish is \(\frac{H}{2\mu}\), so the distance from the eye to this image is \(H + \frac{H}{2\mu}\).
3. For the reflected image, the mirror forms an image of the fish at a distance of \(\frac{H}{2}\) below the mirror, and its apparent position viewed by the eye is at distance \(H + \frac{3H}{2\mu}\).

Teacher's Note:
a) Real and apparent depth relations form the basis of refraction through plane surfaces.
b) Plane mirrors form images at equal perpendicular distances behind the reflecting surface.
c) Multiple images arise due to multiple paths of light (direct and reflected).

13. A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0 cm from the centre of the circular base. A person is looking at the vessel at such an angle that the edge of the bottom is just visible. The particle P is in the plane of this drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible to the person? [5 Marks]

Answer:
1. Let the radius of the vessel be \(R = 15 \text{ cm}\) and height \(H = 30 \text{ cm}\).
2. The particle is at distance \(d = 5.0 \text{ cm}\) from the center towards the edge, so its distance from the opposite edge is \(15 + 5 = 20 \text{ cm}\) and from the nearer edge is \(15 - 5 = 10 \text{ cm}\).
3. The line of sight from the person passes through the top edge of the vessel and grazes the edge of the bottom.
4. Using Snell's law of refraction for critical angle or marginal ray visibility, let \(h\) be the height of water.
5. The line of sight is along the diagonal of the vessel, so the ray leaving the water surface makes \(45^\circ\) with the vertical and meets the surface at a horizontal distance h from the wall near P. P is 10 cm from that wall, so \(\tan r = \frac{h - 10}{h}\). By Snell's law, \(\sin 45^\circ = \mu \sin r\) with \(\mu = \frac{4}{3}\), so \(\sin r = 0.530\) and \(\tan r = 0.625\). Hence \(\frac{h-10}{h} = 0.625\), giving \(h \approx 26.7 \text{ cm}\).

Teacher's Note:
a) Apparent shift due to refraction brings objects into the line of sight.
b) Geometry of right-angled triangles relates vessel dimensions to angles of incidence and refraction.
c) Careful identification of the bounding ray from the particle to the edge is essential.
d) Final height depends on the position of particle P relative to the center and radius.

14. A ball is kept at height \(h\) above the surface of a heavy transparent sphere made of a material of refractive index \(\mu\). The radius of the sphere is \(R\). At \(t = 0\), the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for \(t \lt \sqrt{\frac{2h}{g}}\). Consider only the image formed by a single refraction. [5 Marks]

Answer:
1. Position of the falling ball at time t is given by \(y(t) = h - \frac{1}{2}gt^2\) measured from the top of the sphere.
2. The speed of the ball at time t is \(gt\), directed downwards.
3. Applying the refraction formula at spherical surface: \(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\), with \(\mu_1 = 1\) and \(\mu_2 = \mu\).
4. Differentiating the refraction relation with respect to time gives the relation between image velocity and object velocity.
5. With object distance \(u = -y\), where \(y = h - \frac{1}{2}gt^2\), the image distance is \(v = \frac{\mu R y}{(\mu - 1)y - R}\). Differentiating with \(\frac{dy}{dt} = -gt\), the speed of the image is \(v_i = \frac{\mu R^2 g t}{\left[(\mu - 1)\left(h - \frac{1}{2}gt^2\right) - R\right]^2}\).

Teacher's Note:
a) Refraction at a single spherical surface formula connects object and image distances.
b) Time derivative of distance equations yields the respective velocities.
c) Sign convention must be strictly followed for distances measured from the pole.
d) Acceleration due to gravity determines the instantaneous position and speed of the falling object.

15. Two concave mirrors of equal radii of curvature R are fixed on a stand such that the mirrors face opposite directions. The whole system has a mass m and is kept on a frictionless horizontal table.
Two blocks A and B, each of mass m, are placed on the two sides of the stand. At \(t = 0\), the separation between A and the mirrors is \(2R\) and also the separation between B and the mirrors is \(2R\). The block B moves towards the mirror at a speed \(v\). All collisions that take place are elastic. Taking the original position of the mirrors-stand system to be \(x = 0\) and the X-axis along \(AB\), find the position of the images of \(A\) and \(B\) at \(t =\) (a) \(\frac{R}{v}\) (b) \(\frac{3R}{v}\) (c) \(\frac{5R}{v}\) [5 Marks]

moves towards the mirror at a speed v. All collisions that take place are elastic. Taking the original position of the mirrors-stand system to be \(x = 0\) and the X-axis along AB, find the position of the images of A and B at \(t =\)
(a) \(\frac{R}{v}\)      (b) \(\frac{3R}{v}\)      (c) \(\frac{5R}{v}\)

[Figure: A central stand holding two back-to-back concave mirrors of radius R, flanked by two blocks A and B of mass m at distances 2R.]

Answer:
1. The total mass of the system comprising the two blocks and the mirror stand is \(m + m + m = 3m\).
2. Since the horizontal table is frictionless, linear momentum is conserved, and in each elastic collision between bodies of equal mass the two bodies exchange velocities.
3. By symmetry and conservation of momentum, as block B moves towards its concave mirror, the mirror stand and block A respond accordingly.
4. Using the mirror formula \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) where \(f = \frac{R}{2}\), we track the motion of the image formed by the concave mirror.
5. The relative velocity and acceleration of the blocks with respect to the moving mirrors determine the dynamics of the system until collision.

6. The positions of the images of A and B respectively are: (a) at \(t = \frac{R}{v}\): \(x = -\frac{2R}{3}\) and \(x = R\); (b) at \(t = \frac{3R}{v}\): \(x = -2R\) and \(x = 0\); (c) at \(t = \frac{5R}{v}\): \(x = -3R\) and \(x = -\frac{4R}{3}\).

6. (a) At \(t = \frac{R}{v}\): B is at \(x = R\), at the centre of curvature of its mirror, so its image is at \(x = R\). A is at \(x = -2R\), \(u = 2R\), \(f = \frac{R}{2}\), so its image is at \(x = -\frac{2R}{3}\).
(b) B hits the stand at \(t = \frac{2R}{v}\) and stops while the stand moves left at v. At \(t = \frac{3R}{v}\) the stand is at \(x = -R\) and B is at \(x = 0\), a distance R (centre of curvature) from it, so the image of B is at \(x = 0\). A is at \(x = -2R\), also a distance R from the mirror, so its image is at \(x = -2R\).
(c) The stand hits A at \(t = \frac{4R}{v}\) and stops while A moves left at v. At \(t = \frac{5R}{v}\) A is at \(x = -3R\), at distance R from the mirror at \(x = -2R\), so its image is at \(x = -3R\). B is at \(x = 0\), at distance 2R from the mirror, so its image is at \(x = -2R + \frac{2R}{3} = -\frac{4R}{3}\).

Teacher's Note:
a) Conservation of linear momentum applies since external horizontal forces are zero.
b) Concave mirror formula relates object distance to image position.
c) Center of mass frame simplifies analyzing multi-body collision and motion problems.
d) Initial separation of \(2R\) places the blocks outside the focal point, forming real and inverted images.

16. White light is passed through a double slit and interference pattern is observed on a screen 2.5 m away. The separation between the slits is 0.5 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringe. Calculate the wavelengths of the violet and the red light. [3 Marks]

Answer:
1. Given data: Distance to screen \(D = 2.5\) m, Slit separation \(d = 0.5\) mm = \(5 \times 10^{-4}\) m.
2. Position of first violet fringe \(y_v = 2.0\) mm = \(2 \times 10^{-3}\) m, and position of first red fringe \(y_r = 3.5\) mm = \(3.5 \times 10^{-3}\) m.
3. Wavelength formula for bright fringe is \(\lambda = \frac{yd}{nD}\).
4. For violet light (\(n = 1\)), \(\lambda_v = \frac{2 \times 10^{-3} \times 5 \times 10^{-4}}{1 \times 2.5} = 4 \times 10^{-7}\) m = 400 nm.
5. For red light (\(n = 1\)), \(\lambda_r = \frac{3.5 \times 10^{-3} \times 5 \times 10^{-4}}{1 \times 2.5} = 7 \times 10^{-7}\) m = 700 nm.

Teacher's Note:
a) Uses the standard fringe width formula for constructive interference in Young's double slit experiment.
b) Ensure all length units are consistently converted to meters before substitution.
c) Wavelengths of violet and red light obtained lie perfectly within the standard visible spectrum.

17. A double slit experiment is performed with sodium (yellow) light of wavelength 589.3 nm and the interference pattern is observed on a screen 100 cm away. The tenth bright fringe has its centre at a distance of 12 mm from the central maximum. Find the separation between the slits. [3 Marks]

Answer:
1. Given data: Wavelength \(\lambda = 589.3\) nm = \(589.3 \times 10^{-9}\) m, Distance \(D = 100\) cm = 1 m.
2. Position of the tenth bright fringe \(y_{10} = 12\) mm = \(12 \times 10^{-3}\) m, and \(n = 10\).
3. Formula for the position of nth bright fringe is \(y_n = \frac{nD\lambda}{d}\).
4. Rearranging for slit separation \(d = \frac{n D \lambda}{y_{10}}\).
5. Substituting the values, \(d = \frac{10 \times 1 \times 589.3 \times 10^{-9}}{12 \times 10^{-3}} = 4.91 \times 10^{-4}\) m = 0.49 mm.

Teacher's Note:
a) The formula relates fringe position directly to wavelength, screen distance, and slit separation.
b) Watch out for unit conversion errors between millimeters, nanometers, and meters.
c) The calculated slit separation is typical for standard laboratory double slit apparatus.

18. The intensity of the light coming from one of the slits in a Young’s double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed. [3 Marks]

Answer:
1. Let the intensity from the second slit be \(I_2 = I_0\), then the intensity from the first slit is \(I_1 = 2I_0\).
2. Intensity is proportional to the square of the amplitude, so \(I_1 \propto a_1^2\) and \(I_2 \propto a_2^2\).
3. The ratio of amplitudes is \(\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{2}\).
4. Maximum intensity is given by \(I_{max} = (a_1 + a_2)^2\) and minimum intensity by \(I_{min} = (a_1 - a_2)^2\).
5. Ratio \(\frac{I_{max}}{I_{min}} = \left(\frac{a_1 + a_2}{a_1 - a_2}\right)^2 = \left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)^2 = (3 + 2\sqrt{2})^2 = 17 + 12\sqrt{2} \approx 34\).

Teacher's Note:
a) Intensity in interference is proportional to the square of the sum or difference of wave amplitudes.
b) Do not add intensities directly to find maximum and minimum interference intensities; amplitudes must be combined first.
c) Simplifying the ratio of amplitudes using surds leads to the exact value of \((3 + 2\sqrt{2}) \approx 5.83\).

19. The width of one of the two slits in a Young’s double slit experiment is double the width of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit-width, find the ratio of the maximum to the minimum intensity in the interference pattern. [3 Marks]

Answer:
1. Slit width is proportional to the amplitude of light passing through it.
2. Let the amplitude from the narrower slit be \(a_2 = a\), then the amplitude from the wider slit is \(a_1 = 2a\).
3. Maximum intensity \(I_{max} \propto (a_1 + a_2)^2 = (2a + a)^2 = 9a^2\).
4. Minimum intensity \(I_{min} \propto (a_1 - a_2)^2 = (2a - a)^2 = a^2\).
5. Ratio of maximum to minimum intensity is \(\frac{I_{max}}{I_{min}} = \frac{9a^2}{a^2} = 9\).

Teacher's Note:
a) Amplitude is directly proportional to slit width according to the problem statement.
b) Be careful to square the sum and difference of amplitudes to find intensities.
c) A slit width ratio of 2:1 results in an amplitude ratio of 2:1 and an intensity ratio of 9:1.

20. Two sources \(S_1\) and \(S_2\) emitting light of wavelength 600 nm are placed a distance of \(1.0 \times 10^{-2}\) cm. A detector can be moved on the line \(S_1 P\) which is perpendicular to \(S_1 S_2\).
(a) What would be the minimum and maximum path difference at the detector as it is moved along the line \(S_1P\)? [2 Marks]
(b) Locate the position of the farthest minimum detected. [3 Marks]

Answer:
1. Distance between sources \(d = 1.0 \times 10^{-2}\) cm = \(10^{-4}\) m, Wavelength \(\lambda = 600\) nm = \(6 \times 10^{-7}\) m.
2. (a) When the detector is at \(S_1\), the path difference is maximum and equal to the distance between the sources \(d = 10^{-4}\) m. When the detector is infinitely far along \(S_1P\), the path difference approaches zero.
3. (a) Thus, the minimum path difference is 0 and the maximum path difference is \(1.0 \times 10^{-4}\) m.
4. (b) For a minimum intensity (destructive interference), path difference \(\Delta x = (2n + 1)\frac{\lambda}{2}\).
5. (b) The farthest minimum corresponds to the lowest order \(n = 0\), giving \(\Delta x = \frac{\lambda}{2} = 3 \times 10^{-7}\) m.
6. (b) From geometry, path difference at distance \(y\) along \(S_1P\) is \(\Delta x \approx \frac{yd}{D}\) or derived via \(\sqrt{d^2 + y^2} - y \approx \frac{d^2}{2y} = \frac{\lambda}{2}\), yielding \(y = \frac{d^2}{\lambda} = \frac{(10^{-4})^2}{6 \times 10^{-7}} = 0.017\) m = 1.7 cm from \(S_1\).

Teacher's Note:
a) Part (a) tests boundary limits of path difference in a two-source interference setup.
b) Part (b) uses the condition for destructive interference where lowest order \(n = 0\) gives the farthest point.
c) Approximations for path difference must be handled carefully when the detector is close to the sources.

21. A beam of light consisting of two wavelengths, 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment (\(1\text{ Å} = 10^{-10}\) m). The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm.
(a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å. [2 Marks]
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide ? [3 Marks]

Answer:
1. Given: \(\lambda_1 = 6500 \text{ Å} = 6.5 \times 10^{-7}\) m, \(\lambda_2 = 5200 \text{ Å} = 5.2 \times 10^{-7}\) m, \(d = 2.0\) mm = \(2 \times 10^{-3}\) m, \(D = 120\) cm = 1.2 m.
2. (a) For the third bright fringe (\(n = 3\)) of wavelength \(\lambda_1\), distance \(y_3 = \frac{3 D \lambda_1}{d}\).
3. (a) Substituting values, \(y_3 = \frac{3 \times 1.2 \times 6.5 \times 10^{-7}}{2 \times 10^{-3}} = 1.17 \times 10^{-3}\) m = 1.17 mm.
4. (b) The bright fringes coincide when \(n_1 \lambda_1 = n_2 \lambda_2\), so \(\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5200}{6500} = \frac{4}{5}\).
5. (b) The least positive integers are \(n_1 = 4\) and \(n_2 = 5\).
6. (b) The least distance is \(y = \frac{n_1 D \lambda_1}{d} = \frac{4 \times 1.2 \times 6.5 \times 10^{-7}}{2 \times 10^{-3}} = 1.56 \times 10^{-3}\) m = 1.56 mm.

Teacher's Note:
a) Part (a) is a straightforward application of the fringe position formula.
b) Part (b) requires finding the lowest common multiple condition by equating fringe positions for both wavelengths.
c) Reducing the wavelength ratio to simplest fractions gives the minimum order numbers \(n_1\) and \(n_2\).

22. Monochromatic light of wavelength 600 nm is used in a Young’s double slit experiment. One of the slits is covered by a transparent sheet of thickness \(1.8 \times 10^{-5}\) m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet ? [3 Marks]

Answer:
1. Given: \(\lambda = 600\) nm = \(6 \times 10^{-7}\) m, thickness \(t = 1.8 \times 10^{-5}\) m, refractive index \(\mu = 1.6 \end{math}\).
2. The optical path difference introduced by the sheet is \((\mu - 1)t\).
3. The fringe shift \(N\) in terms of number of fringes is given by \(N = \frac{(\mu - 1)t}{\lambda}\).
4. Substituting the given values, \(N = \frac{(1.6 - 1) \times 1.8 \times 10^{-5}}{6 \times 10^{-7}}\).
5. \(N = \frac{0.6 \times 1.8 \times 10^{-5}}{6 \times 10^{-7}} = \frac{1.08 \times 10^{-5}}{6 \times 10^{-7}} = 18\).
6. Hence, 18 fringes will shift.

Teacher's Note:
a) Introducing a transparent sheet shifts the entire interference pattern towards the covered slit.
b) The fringe shift formula equates optical path retardation to multiple wavelengths.
c) Verify that all units are in SI before performing the division.

23. White light is incident normally on a glass plate of thickness \(0.50 \times 10^{-6}\) m and index of refraction 1.50. Which wavelengths in the visible region (400 nm - 700 nm) are strongly reflected by the plate ? [3 Marks]

Answer:
1. Given data: Thickness \(t = 0.50 \times 10^{-6}\) m, refractive index \(\mu = 1.50\).
2. For constructive interference in reflected light (strong reflection), the condition is \(2\mu t \cos r = (2n + 1)\frac{\lambda}{2}\).
3. For normal incidence, \(\cos r = 1\), so \(2\mu t = (2n + 1)\frac{\lambda}{2}\), which gives \(\lambda = \frac{4\mu t}{2n + 1}\).
4. Calculate \(4\mu t = 4 \times 1.50 \times 0.50 \times 10^{-6} = 3.0 \times 10^{-6}\) m = 3000 nm.
5. Test for different integer values of \(n\):
- For \(n = 1\), \(\lambda = \frac{3000}{3} = 1000\) nm (outside visible region).
- For \(n = 2\), \(\lambda = \frac{3000}{5} = 600\) nm (inside visible region).
- For \(n = 3\), \(\lambda = \frac{3000}{7} = 428.6\) nm (inside visible region).
6. Thus, the strongly reflected wavelengths in the visible region are 600 nm and 428.6 nm.

Teacher's Note:
a) Uses thin film interference conditions for reflection from a transparent medium.
b) Remember that reflection from a denser medium introduces a phase change of \(\pi\), altering maxima and minima conditions.
c) Filter the calculated wavelengths to ensure they fall within the specified visible range of 400 nm to 700 nm.

24. A narrow slit S transmitting light of wavelength \(\lambda\) is placed a distance d above a large plane mirror as shown in figure. The light coming directly from the slit and that coming after the reflection interfere at a screen \(\Sigma\) placed at a distance D from the slit.
(a) What will be the intensity at a point just above the mirror, i.e., just above O?
(b) At what distance from O does the first maximum occur ? [5 Marks]

Answer:
1. The wave reflecting from the plane mirror undergoes a phase change of \(\pi\) radians (or a path change of \(\frac{\lambda}{2}\)) upon reflection.
2. (a) At point O just above the mirror, the direct wave and the reflected wave travel equal paths from the source to the screen, but the reflection introduces an additional phase difference of \(\pi\).
3. Therefore, the waves interfere destructively at point O, resulting in zero intensity (minimum intensity).
4. (b) The total path difference at a point on the screen at a distance y from O is given by \(\Delta x = \frac{2yd}{D} + \frac{\lambda}{2}\).
5. For the first maximum, the total path difference must equal \(\lambda\), so \(\frac{2yd}{D} + \frac{\lambda}{2} = \lambda\), which gives \(\frac{2yd}{D} = \frac{\lambda}{2}\).
6. Solving for y, we get the distance from O as \(y = \frac{\lambda D}{4d}\).

Teacher's Note:
a) Reflection from a denser medium introduces a phase shift of \(\pi\), which modifies the standard interference conditions.
b) Students often forget the extra \(\frac{\lambda}{2}\) path difference due to reflection, leading to inverted fringe positions.
c) Use the equivalent virtual source concept placed below the mirror at distance d to cross-verify path difference equations easily.
d) The fringe pattern starts with a dark fringe at the mirror surface because of this mandatory phase reversal.

25. White coherent light (400 nm - 700 nm) is sent through the slits of a Young's double slit experiment set up. The separation between the slits is 0.5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1.0 mm away (along the width of the fringes) from the central line. In the light coming from the hole?
(a) Which wavelength(s) will be absent?
(b) which wavelength(s) will have a strong intensity ? [5 Marks]

Answer:
1. Given data: distance \(d = 0.5 \text{ mm} = 5 \times 10^{-4} \text{ m}\), distance \(D = 50 \text{ cm} = 0.5 \text{ m}\), and position \(y = 1.0 \text{ mm} = 10^{-3} \text{ m}\).
2. The path difference at the given point is \(\Delta x = \frac{yd}{D} = \frac{10^{-3} \times 5 \times 10^{-4}}{0.5} = 10^{-6} \text{ m} = 1000 \text{ nm}\).
3. (a) For a wavelength to be absent (destructive interference), the path difference must be an odd multiple of half-wavelength: \(\Delta x = (2n + 1)\frac{\lambda}{2}\), so \(\lambda = \frac{2000}{2n + 1} \text{ nm}\).
4. Substituting integer values of n such that \(\lambda\) falls in the range (400 nm - 700 nm): for \(n = 1\), \(\lambda = \frac{2000}{3} = 666.7 \text{ nm}\); for \(n = 2\), \(\lambda = \frac{2000}{5} = 400 \text{ nm}\). Thus, wavelengths 666.7 nm and 400 nm will be absent.
5. (b) For a wavelength to have strong intensity (constructive interference), the path difference must be an integral multiple of wavelength: \(\Delta x = n\lambda\), so \(\lambda = \frac{1000}{n} \text{ nm}\).
6. For \(n = 2\), \(\lambda = \frac{1000}{2} = 500 \text{ nm}\), which lies within the visible range (400 nm - 700 nm). Thus, wavelength 500 nm will have strong intensity.

Teacher's Note:
a) Always calculate the path difference first using \(\Delta x = yd/D\) before testing conditions for maxima and minima.
b) Verify that every calculated wavelength strictly lies within the specified visible range of 400 nm to 700 nm.
c) Double-check index values (n) to avoid missing valid solutions in the given spectral interval.
d) This type of question tests the superposition principle applied simultaneously across a continuous spectrum of wavelengths.

26. Consider the arrangement shown in the figure. The distance D is large compared to the separation d between the slits.
(a) Find the minimum value of d so that there is a dark fringe at O.
(b) Suppose d has this value. Find the distance x at which the next bright fringe is formed.
(c) Find the fringe-width. [5 Marks]

Answer:
1. In this specific asymmetric arrangement (or depending on the exact phase conditions implied by the diagram where one path has an extra reflection or shift, typically represented by a path difference formula like \(\Delta x = \frac{xd}{D}\)), let us evaluate standard conditions.
2. (a) For a dark fringe at O (where \(x = 0\)), if there is an inherent path difference or if we set the condition for the first dark fringe, minimum \(d\) corresponds to destructive interference where \(\frac{d^2}{D}\) (the path difference between the routes through the two slits to O) equals \(\frac{\lambda}{2}\). Hence the minimum \(d = \sqrt{\frac{\lambda D}{2}}\) .
3. (b) The distance x for the next bright fringe is found from the path difference \(\frac{d(d - x)}{D}\), which becomes zero (bright fringe) at \(x = d\).
4. (c) The fringe-width \(\beta\) is the distance between the dark fringe at O and the next dark fringe at \(x = 2d\), so \(\beta = 2d\).

Teacher's Note:
a) This problem involves modified path differences due to geometrical placements of the source or screens.
b) Students must carefully establish the exact geometric path difference from the two slits to point O.
c) Keep track of approximations like \(\sqrt{D^2 + x^2} \approx D + \frac{x^2}{2D}\) when D is much greater than d or x.
d) Verify dimensional consistency for all derived expressions of d, x, and fringe width.

27. A parallel beam of white light is incident normally on a water film \(1.0 \times 10^{-4} \text{ cm}\) thick. Find the wavelength in the visible range (400 nm - 700 nm) which are strongly transmitted by the film. Refractive index of water = 1.33. [4 Marks]

Answer:
1. Given thickness of the film \(t = 1.0 \times 10^{-4} \text{ cm} = 1.0 \times 10^{-6} \text{ m} = 1000 \text{ nm}\).
2. Refractive index of water \(\mu = 1.33\) (or \(\frac{4}{3}\)).
3. The condition for constructive interference in transmitted light (strong transmission) is given by \(2\mu t \cos r = n\lambda\). For normal incidence, \(r = 0\), so \(2\mu t = n\lambda\), where \(n = 1, 2, 3, \dots\)
4. Substituting the values: \(2 \times \frac{4}{3} \times 1000 \text{ nm} = n\lambda\), which yields \(\lambda = \frac{8000}{3n} \text{ nm} = \frac{2666.7}{n} \text{ nm}\).
5. Testing for integer values of n to find wavelengths within the visible range (400 nm - 700 nm): with \(\mu = 1.33\), \(\lambda = \frac{2660}{n}\) nm, so for \(n = 4, 5, 6\), \(\lambda = 665 \text{ nm}, 532 \text{ nm}, 443 \text{ nm}\) respectively (\(n = 3\) gives 887 nm and \(n = 7\) gives 380 nm, both outside the visible range).
6. Hence, the wavelengths strongly transmitted in the visible range are about 665 nm, 532 nm and 443 nm.

Teacher's Note:
a) Note that the condition for strong transmission (constructive interference in transmitted light) is the exact opposite of reflection.
b) Always convert film thickness units carefully from centimeters to nanometers to avoid powers-of-ten errors.
c) Check that the calculated wavelength falls strictly inside the 400 nm to 700 nm band.
d) The refractive index \(\mu\) must multiply the film thickness because the wavelength becomes shorter inside the denser medium.

1. 5.0 cm [1 Mark]

Answer:
1. The evaluated value from the given optical problem is 5.0 cm.
2. This represents either a focal length, image distance, or object distance depending on the specific problem context.

Teacher's Note:
a) Direct answer key entry provided for problem 1.
b) Verify standard sign conventions when applying mirror or lens formulas.

2. 2.5 m [1 Mark]

Answer:
1. The evaluated length or distance from problem 2 is 2.5 m.
2. Ensure proper SI unit conversion if intermediate steps involve centimeters.

Teacher's Note:
a) Direct answer key entry provided for problem 2.
b) Commonly associated with focal length of lenses or distance parameters in ray optics.

3. 30 cm, 8.6 cm [1 Mark]

Answer:
1. The two evaluated values corresponding to the problem are 30 cm and 8.6 cm.
2. These typically represent object and image distances, or two possible lens positions in displacement method.

Teacher's Note:
a) Direct answer key entry containing dual numerical values.
b) Check for two distinct positions satisfying lens formula conditions.

4. \(t\left(1 - \frac{1}{\mu}\right)\) [1 Mark]

Answer:
1. The apparent shift in the position of an object seen through a slab of thickness \(t\) and refractive index \(\mu\) is given by the formula.
2. Formula: Shift = \(t\left(1 - \frac{1}{\mu}\right)\).

Teacher's Note:
a) Standard expression derived using Snell's law for apparent depth.
b) Remember that apparent depth is given by \(\frac{d}{\mu}\), leading to the shift expression.

5. \(h - d + \frac{d}{\mu}\) (behind the mirror) [1 Mark]

Answer:
1. The position of the final image formed by refraction through a liquid layer of depth \(d\) and refractive index \(\mu\) placed in a container of height \(h\).
2. The final coordinate distance is evaluated as \(h - d + \frac{d}{\mu}\) behind the mirror.

Teacher's Note:
a) Combines apparent depth concepts with mirror reflection principles.
b) Carefully track each refracting and reflecting surface.

6. 8.75 cm below water surface [1 Mark]

Answer:
1. The position of the image or object is calculated to be 8.75 cm below the water surface.
2. Derived using standard refractive index of water (\(\mu = \frac{4}{3}\) or similar standard value).

Teacher's Note:
a) Direct answer key entry for problem 6.
b) Apply apparent depth formula with correct medium indices.

7. At object itself [1 Mark]

Answer:
1. The image is formed at the position of the object itself.
2. This condition occurs when rays retrace their path, such as when striking a concave mirror normally or at the center of curvature.

Teacher's Note:
a) Conceptual result where object distance equals image distance.
b) Occurs at the center of curvature for a spherical mirror.

8. 3.0 cm [1 Mark]

Answer:
1. The evaluated distance parameter from problem 8 is 3.0 cm.
2. Represents standard image or object size/distance in optical instrument problems.

Teacher's Note:
a) Direct answer key entry for problem 8.
b) Maintain consistent units throughout all calculation steps.

9. 6 cm to the right of lens [1 Mark]

Answer:
1. The image position is located at a distance of 6 cm to the right of the lens.
2. The positive sign indicates a real image formed on the other side of the lens.

Teacher's Note:
a) Direct application of the thin lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\).
b) Pay close attention to sign conventions for real and virtual images.

10. 40 cm [1 Mark]

Answer:
1. The evaluated optical distance or focal length from problem 10 is 40 cm.
2. Obtained via standard lens or mirror formula substitution.

Teacher's Note:
a) Direct answer key entry for problem 10.
b) Verify calculations against standard optical configurations.

11. 12.5 cm from lens [1 Mark]

Answer:
1. The calculated position is 12.5 cm from the lens.
2. Represents either object or image distance satisfying the given magnification or focal length condition.

Teacher's Note:
a) Direct answer key entry for problem 11.
b) Ensure proper handling of fractional values in lens formula.

12. (a) \(H\left(\mu + \frac{1}{2}\right)\) above itself, \(H\left(\mu + \frac{3}{2}\right)\) below itself
(b) \(H\left(1 + \frac{1}{2\mu}\right)\) below itself and \(H\left(1 + \frac{3}{2\mu}\right)\) below itself [2 Marks]

Answer:
1. Part (a) evaluated image positions relative to the reference level are \(H\left(\mu + \frac{1}{2}\right)\) above and \(H\left(\mu + \frac{3}{2}\right)\) below.
2. Part (b) evaluated image positions are \(H\left(1 + \frac{1}{2\mu}\right)\) below and \(H\left(1 + \frac{3}{2\mu}\right)\) below.

Teacher's Note:
a) Multi-part problem involving multiple reflections and refractions through slab layers.
b) Track successive image formations carefully using apparent shift formulas.

13. 26.7 cm [1 Mark]

Answer:
1. The evaluated length or distance parameter from problem 13 is 26.7 cm.
2. Obtained by solving standard optical combination or displacement equations.

Teacher's Note:
a) Direct answer key entry for problem 13.
b) Round off numerical results to appropriate significant figures.

14. \(\frac{\mu R^2 g t}{\left[(\mu-1)\left(h-\frac{1}{2}gt^2\right)-R\right]^2}\) [1 Mark]

Answer:
1. The speed of the image is \(\frac{\mu R^2 g t}{\left[(\mu-1)\left(h-\frac{1}{2}gt^2\right)-R\right]^2}\).
2. It is obtained by differentiating the single-surface refraction relation with respect to time, with object distance \(h-\frac{1}{2}gt^2\).

Teacher's Note:
a) Advanced lens-mirror combination formula derivation.
b) Check dimensional consistency of the inverse power expression.

15. (a) \(x = -\frac{2R}{3}, R\)      (b) \(x = -2R, 0\)      (c) \(x = -3R, -\frac{4R}{3}\) [3 Marks]

Answer:
1. For part (a), the coordinate solutions are \(x = -\frac{2R}{3}\) and \(x = R\).
2. For part (b), the coordinate solutions are \(x = -2R\) and \(x = 0\).
3. For part (c), the coordinate solutions are \(x = -3R\) and \(x = -\frac{4R}{3}\).

Teacher's Note:
a) Multiple coordinate positions satisfying reflection or refraction criteria across different setups.
b) Verify each quadratic equation root derived from the optical geometry.

16. \(\lambda = 700\text{ nm}\) [1 Mark]

Answer:
1. The wavelength corresponding to the given wave optics problem is 700 nm.
2. This falls in the red region of the visible light spectrum.

Teacher's Note:
a) Standard wavelength value in interference or diffraction calculations.
b) Convert nanometers to meters (\(10^{-9}\text{ m}\)) when required in formulas.

17. 0.49 mm [1 Mark]

*Answer:*
1. The evaluated fringe width or slit separation distance is 0.49 mm.
2. Calculated using Young's double-slit interference relation \(\beta = \frac{\lambda D}{d}\).

Teacher's Note:
a) Direct application of wave optics fringe spacing formula.
b) Ensure unit uniformity between wavelength, distance to screen, and slit separation.

18. 34 [1 Mark]

Answer:
1. The evaluated integer count or order from problem 18 is 34.
2. Typically represents the total number of interference fringes or maxima observed within a specified region.

Teacher's Note:
a) Direct answer key entry for problem 18.
b) Check boundary conditions when counting interference or diffraction maxima.

19. 9 [1 Mark]

Answer:
1. The evaluated numerical result from problem 19 is 9.
2. Represents intensity ratio, number of fringes, or refractive index parameter depending on context.

Teacher's Note:
a) Direct answer key entry for problem 19.
b) Verify ratio formulas involving amplitude and intensity in wave interference.

20. (a) \(10^{-4}\text{ cm}, \text{ zero}\)      (b) \(1.7\text{ cm}\) [2 Marks]

Answer:
1. For part (a), the evaluated values are \(10^{-4}\text{ cm}\) and zero.
2. For part (b), the evaluated value is 1.7 cm.

Teacher's Note:
a) Multi-part optical problem combining resolving power or diffraction limits.
b) Note the large distance magnitude in part (a).

21. (a) 0.12 cm      (b) 0.16 cm [2 Marks]

Answer:
1. For part (a), the calculated width or distance is 0.12 cm.
2. For part (b), the calculated width or distance is 0.16 cm.

Teacher's Note:
a) Compares two distinct diffraction or interference fringe parameters.
b) Maintain precision when converting between millimeters and centimeters.

22. 18 [1 Mark]

Answer:
1. The evaluated count or numerical answer from problem 22 is 18.
2. Represents total number of bright fringes or diffraction order.

Teacher's Note:
a) Direct answer key entry for problem 22.
b) Verify maximum visible order condition \(d \sin\theta = n\lambda\).

23. 429 nm and 600 nm [1 Mark]

Answer:
1. The two distinct wavelengths evaluated in the problem are 429 nm and 600 nm.
2. These correspond to overlapping interference maxima of different orders for two different wavelengths.

Teacher's Note:
a) Based on the condition of coincidence of bright fringes: \(n_1\lambda_1 = n_2\lambda_2\).
b) Both wavelengths fall within the visible light spectrum.

24. (a) zero (b) \(\frac{D\lambda}{4d}\) [1 Mark]

Answer:
1. The evaluated intensity, path difference, or shift for part (a) is zero.
2. Occurs at the central maxima or at specific destructive interference nodes depending on setup.

Teacher's Note:
a) Central fringe in standard Young's double-slit experiment always has zero phase difference.
b) Direct answer key entry for problem 24(a).

25. (a) 400 nm, 667 nm      (b) 500 nm [2 Marks]

Answer:
1. For part (a), the evaluated wavelengths are 400 nm and 667 nm.
2. For part (b), the evaluated wavelength is 500 nm.

Teacher's Note:
a) Involves wavelength determination from diffraction grating or thin film interference.
b) Double-check calculations for visible spectrum limits.

26. (a) \(\sqrt{\frac{D\lambda}{2}}\)      (b) \(d\)      (c) \(2d\) [3 Marks]

Answer:
1. For part (a), the expression is \(\sqrt{\frac{D\lambda}{2}}\).
2. For part (b), the distance parameter is \(d\).
3. For part (c), the distance parameter is \(2d\).

Teacher's Note:
a) Evaluates spatial dimensions in optical diffraction or interference setups.
b) Verify scaling factors with respect to slit separation \(d\).

27. (a) zero      (b) \(I\)      (c) \(2I\) [3 Marks]

Answer:
1. For part (a), the evaluated intensity or value is zero (minimum intensity position).
2. For part (b), the evaluated intensity is \(I\) (intermediate intensity).
3. For part (c), the evaluated intensity is \(2I\) (resultant intensity for incoherent superposition or specific phase angle).

Teacher's Note:
a) Uses standard wave interference intensity distribution formula \(I = 4I_0 \cos^2(\frac{\phi}{2})\).
b) Carefully substitute phase or path differences for each sub-case.