CBSE Class 12 Physics Optics Advanced Problems

CBSE Class 12 Physics Optics Advanced 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.

ADVANCED PROBLEMS
CLASS 12 PHYSICS : OPTICS

1. A 16 cm long pencil is placed at 45º angle with its centre 15.0 cm above the optic axis and 45.0 cm from a convex lens of focal length 20 cm as shown in figure. Assume that the diameter of the lens is large enough for the paraxial approximation to be valid :
(a) where is the image of the pencil formed. (Give the location of the images of the points A, B and C w.r.t. coordinate system shown in the figure)
(b) what is the length of the image (the distance between the images of points A and B).

Answer:
1. Take the lens at the origin, with x measured along the axis and y upwards. The object points are C (-45, 15), A (-45 - 8cos45°, 15 - 8sin45°) = (-50.66, 9.34) and B (-39.34, 20.66), all in cm.
2. (a) For each point use 1/v = 1/f + 1/u with f = 20 cm, and the image height y' = (v/u) y.
3. C: u = -45, so 1/v = 1/20 - 1/45 and v = 36 cm; magnification v/u = -0.8, so y' = -12 cm. Image of C = (36, -12) cm.
4. A: u = -50.66, so v = 33.0 cm; magnification = -0.652, so y' = -6.1 cm. Image of A = (33.0, -6.1) cm.
5. B: u = -39.34, so v = 40.7 cm; magnification = -1.034, so y' = -21.4 cm. Image of B = (40.7, -21.4) cm.
6. (b) Length of the image A'B' = √[(40.7 - 33.0)² + (-21.4 + 6.1)²] = √(7.7² + 15.3²) = 17.1 cm.
7. Hence, the image is real, inverted and inclined to the axis, with A' (33.0, -6.1), C' (36, -12), B' (40.7, -21.4) and length 17.1 cm.

Teacher's Note:
a) The image of the pencil is not parallel to the pencil, because the longitudinal and transverse magnifications are different.
b) The image is longer than the object since the points nearer the focus are magnified more.

2. Two converging lenses of the same focal length f are separated by distance 2f. The axis of the second lens is inclined at angle θ = 60º with respect to the axis of the first lens. A parallel paraxial beam of light is incident from left side of the lens. Find the coordinates of the final image with respect to optical centre of firs lens as origin.

Answer:
1. The parallel beam is focused by the first lens at its focus F, at (f, 0), which is the mid-point between the two lenses. This point acts as the object for the second lens.
2. F is at a distance f from the centre of the second lens, along the line that makes 60° with the axis of the second lens. So the object distance along the axis of lens 2 is u = f cos 60° = f/2, and its height above that axis is h = f sin 60° = (√3/2)f.
3. Since u = f/2 is less than f, the image is virtual: 1/v = 1/f - 1/(f/2) = -1/f, so v = -f, on the same side as the object.
4. The magnification is v/u = 2, so the image height is √3 f.
5. Image position relative to the second lens = (-f) along its axis and √3 f perpendicular to its axis. Taking components in the x and y directions: x = 2f - f cos 60° - √3f sin 60° = 2f - f/2 - 3f/2 = 0 and y = -f sin 60° + √3f cos 60° = 0.
6. Hence, the final image is formed at (0, 0), that is, at the optical centre of the first lens.

Teacher's Note:
a) The point (0, 0) is a virtual image seen through the second lens.
b) Resolve every displacement along the axis of the second lens and perpendicular to it before converting to x and y.

3. A double convex lens forms a real image of an object on a screen which is fixed. Now the lens is given a constat velocity v₀ along its axis and away from the screen. For the purpose of forming the image always on the screen, the object is also required to be given appropriate velocity. Find the velocity of the object at the instant its size is n times the size of the image. (n < 1).

Answer:
1. Let u be the object distance and v the distance of the lens from the screen (the image distance). Then 1/u + 1/v = 1/f, with f constant.
2. Differentiating with respect to time: -(1/u²)(du/dt) - (1/v²)(dv/dt) = 0, so the object-lens distance changes at the rate du/dt = -(u²/v²)(dv/dt), with u and v taken as positive distances.
3. The lens moves away from the screen at v₀, so the image distance increases at the rate dv/dt = v₀.
4. The object size is n times the image size, so the magnification is v/u = 1/n, that is, u/v = n.
5. The object distance changes at the rate -n²v₀ relative to the lens, that is, the object-lens distance decreases. The velocity of the object relative to the screen = v₀ - n²v₀ = (1 - n²) v₀, in the direction of motion of the lens, away from the screen.
6. Hence, the velocity of the object is (1 - n²) v₀ away from the screen.

Teacher's Note:
a) The printed answer is ((1 - n²)/n²) v₀ towards the screen. It corresponds to the image being n times the object (magnification n, with u/v = 1/n), which is not what the question says; for the question as printed the result is (1 - n²) v₀ away from the screen.
b) Velocity relative to the screen = velocity of the lens + rate of change of the object distance.

4. A ray of light is incident on the surface of a sphere of refractive index √7/2. Other half of the sphere is silvered. After refraction it is reflected and then refracted out of the sphere again such that the total deviation is minimum. Find :
(a) the angle of incidence of the ray
(b) total deviation of the ray, (sin 41º = \( \sqrt{\frac{3}{7}} \)).

Answer:
1. Let the angle of incidence be i and the angle of refraction r, with sin i = μ sin r and μ = √7/2.
2. The deviation at the first refraction is (i - r), at the silvered surface it is (180° - 2r) and at the second refraction it is (i - r). Total deviation δ = 2(i - r) + 180° - 2r = 180° + 2i - 4r.
3. For the minimum, dδ/di = 0, so 2 - 4(dr/di) = 0, that is, dr/di = 1/2. From cos i = μ cos r (dr/di), we get cos i = (μ/2) cos r.
4. Squaring and using cos²r = 1 - sin²i/μ²: cos²i = (μ²/4)(1 - sin²i/μ²) = μ²/4 - (sin²i)/4. With cos²i = 1 - sin²i this gives (3/4) sin²i = 1 - μ²/4.
5. Here μ² = 7/4, so μ²/4 = 7/16 and (3/4) sin²i = 9/16, that is, sin²i = 3/4.
6. (a) sin i = √3/2, so i = 60°.
7. (b) sin r = sin 60°/μ = (√3/2)(2/√7) = √(3/7), so r = 41°. δ = 180° + 2(60°) - 4(41°) = 180° + 120° - 164° = 136°.
8. Hence, the angle of incidence is 60° and the total deviation is 136°.

Teacher's Note:
a) Be careful to add the three deviations in the same sense; the reflection at the back surface turns the ray through 180° - 2r.
b) The value of r = 41° follows directly from the data sin 41° = √(3/7) given in the question.

5. The refracting angle of prism is 60º. A ray of light passing through the prism undergoes a deviation of 30º. If the prism is rotated keeping the incident ray fixed through an angle 30º, the same deviation is again obtained. Calculate the refractive index of the prism.

Answer:
1. Deviation δ = i + e - A, so i + e = A + δ = 60° + 30° = 90°, where i and e are the angles of incidence and emergence.
2. The same deviation occurs for two angles of incidence, and these two are interchanged: if the incidence is i, the second incidence is e. Rotating the prism by 30° changes the angle of incidence by 30°, so e - i = 30°.
3. Solving, i = 30° and e = 60°.
4. At the first face sin 30° = μ sin r₁, and at the second face sin 60° = μ sin r₂, with r₁ + r₂ = 60°.
5. Dividing: sin(60° - r₁)/sin r₁ = sin 60°/sin 30° = √3. Expanding, (√3/2) cot r₁ - 1/2 = √3, so cot r₁ = 2 + 1/√3 = 2.577, giving r₁ = 21.2°.
6. μ = sin 30°/sin 21.2° = 0.5/0.3617 = 1.382.
7. Hence, the refractive index of the prism is 1.382.

Teacher's Note:
a) A prism gives the same deviation for two conjugate angles of incidence, i and e, which are the incidence and emergence angles of the same ray path.
b) Check: sin 60°/sin 38.8° = 0.866/0.6266 = 1.382.

7. A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid.

Answer:
1. The glass lens is biconvex, with both radii equal to R. Its focal length: 1/f_g = (3/2 - 1)(1/R + 1/R) = 1/R.
2. The liquid between the lens and the plane mirror forms a plano-concave liquid lens (the curved face is the lower face of the glass lens): 1/f_l = -(μ - 1)/R.
3. The combination of lens and liquid has focal length F, where 1/F = 1/R - (μ - 1)/R = (2 - μ)/R.
4. The object coincides with its image only if the rays fall normally on the plane mirror, that is, they emerge parallel from the combination. So the object is at the focus: object distance = F.
5. With water: 15 = F, so 1/15 = (2 - 4/3)/R = 2/(3R), which gives R = 10 cm.
6. With the other liquid: 25 = F, so 1/25 = (2 - μ)/10, that is, 2 - μ = 0.4.
7. μ = 1.6.
8. Hence, the refractive index of the liquid is 1.6.

Teacher's Note:
a) The rays retrace their path only when they hit the plane mirror along the normal.
b) The liquid lens is concave, so it reduces the power of the glass lens.

8. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. If A'B' is the image after refraction from the lens and the reflection from the mirror, find the distance of A'B' from the pole of the mirror and obtain its magnification. Also locate positions of A' and B' with respect to the optic axis RS.

Answer:
1. Lens: u = -20 cm, f = 15 cm, so 1/v = 1/15 - 1/20 = 1/60 and v = 60 cm. The magnification is -3, so the image is 3.6 cm tall and inverted, formed 60 cm from the lens, that is, 30 cm behind the mirror.
2. This image acts as a virtual object for the mirror, 30 cm behind its pole.
3. Mirror: f = -30 cm and u = +30 cm, so 1/v = 1/f - 1/u = -1/30 - 1/30 = -1/15 and v = -15 cm. The final image is real and lies 15 cm in front of the mirror (between the mirror and the lens).
4. The mirror magnification = -v/u = +0.5. Total magnification = (-3)(0.5) = -3/2, so A'B' is inverted and 1.8 cm long.
5. Heights with respect to RS (PQ is 0.6 cm above RS): the lens image of B is at +0.6 cm and that of A is at 0.6 - 3.6 = -3.0 cm. After the mirror these become +0.6 × 0.5 = +0.3 cm and -3.0 × 0.5 = -1.5 cm.
6. Hence, A'B' is 15 cm from the pole of the mirror with magnification -3/2; B' is 0.3 cm above RS and A' is 1.5 cm below RS.

Teacher's Note:
a) A mirror gets a virtual object when the lens image falls behind it, and then it can form a real image.
b) The vertical offset of the axes shifts the heights by 0.6 cm before the mirror magnification is applied.

9. The X-Y plane is the boundary between two transparent media. Medium-1 with z ≥ 0 has a refractive index √2 and medium-2 with z ≤ 0 has a refractive index √3 . A ray of light in medium-1 given by vector A = 6√3 i + 8√3 j - 10 k is incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium-2.

Answer:
1. The normal to the boundary is along the z-axis. The magnitude |A| = √(108 + 192 + 100) = √400 = 20.
2. cos θ₁ = 10/20 = 1/2, so the angle of incidence θ₁ = 60°.
3. Snell's law: √2 sin 60° = √3 sin θ₂, so sin θ₂ = (√2/√3)(√3/2) = 1/√2 and θ₂ = 45°.
4. The components of the ray parallel to the boundary stay in the same direction. The unit vector of the horizontal part of A is (6√3 i + 8√3 j)/(10√3) = (3i + 4j)/5, and the horizontal part of the refracted unit vector has magnitude sin 45° = 1/√2.
5. The vertical part has magnitude cos 45° = 1/√2, directed along -k.
6. Refracted unit vector = (1/√2)(3i + 4j)/5 - (1/√2)k = (3i + 4j - 5k)/(5√2).
7. Hence, the unit vector along the refracted ray is (3i + 4j - 5k)/(5√2).

Teacher's Note:
a) The tangential direction does not change on refraction, only the angle with the normal changes.
b) Check: (9 + 16 + 25)/50 = 1, so the vector has unit length.

10. A prism of refractive index n₁ and another prism of refractive index n₂ are stuck together with a gap as shown in the figure. The angles of the prism are as shown. n₁ and n₂ depend on λ, the wavelength of light, according to :
\( n_1 = 1.20 + \frac{10.8 \times 10^4}{\lambda^2} \) and
and \( n_2 = 1.45 + \frac{1.80 \times 10^4}{\lambda^2} \) where λ is in nm.
(a) Calculate the wavelength λ₀ for which rays incident at any angle on the interface BC pass through without bending at that interface.
(b) For light of wavelength λ₀, find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum.

Answer:
1. (a) Rays pass through BC without bending for every angle of incidence only if n₁ = n₂.
2. 1.20 + 10.8 × 10⁴/λ² = 1.45 + 1.80 × 10⁴/λ², so 9.0 × 10⁴/λ² = 0.25, that is, λ² = 3.6 × 10⁵ nm² and λ₀ = 600 nm.
3. (b) At λ₀, n₁ = n₂ = 1.20 + 10.8 × 10⁴/(3.6 × 10⁵) = 1.50, so the two prisms act as one prism of index 1.5 whose refracting faces are AC and BD.
4. The angles of the figure are: A = 60°, ∠ABC = 40° and ∠CBD = 20°, so ∠ABD = 60°. The faces AC and BD meet at 60° (triangle formed with AB is equilateral), so the effective prism angle is 60°.
5. For minimum deviation, r₁ = r₂ = 30°. So sin i = 1.5 sin 30° = 0.75 and i = sin⁻¹(3/4) = 48.6°.
6. Hence, λ₀ = 600 nm and the angle of incidence is i = sin⁻¹(3/4) ≈ 48.6°.

Teacher's Note:
a) When n₁ = n₂ the interface BC has no optical effect, so only the outer faces AC and BD matter.
b) At minimum deviation the ray inside the prism is parallel to its base, with r = A/2.

11. A thin equiconvex lens of glass of refractive index μ = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water (μ = 4/3). On the opposite side of the lens, a mirror is placed inside the tank. The tank wall is perpendicular to the lens axis as shown in figure. The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system.

Answer:
1. In air, 1/f = (3/2 - 1)(2/R) = 1/R, so R = 0.3 m for the equiconvex lens.
2. For a lens with air on one side and water on the other, the refraction equation (sum of the two surfaces) is (4/3)/v - 1/u = (3/2 - 1)/R + (4/3 - 3/2)/(-R) = (1/2 + 1/6)/0.3 = 2.22 m⁻¹.
3. First refraction: u = -0.9 m, so (4/3)/v = 2.22 - 1.11 = 1.11, which gives v = 1.2 m. The image would be formed 1.2 m behind the lens, that is, 0.4 m behind the plane mirror (which is at 0.8 m).
4. The plane mirror forms the image of this virtual object 0.4 m in front of it, that is, at 0.8 - 0.4 = 0.4 m from the lens, in the water. The light now travels back towards the lens.
5. Second refraction (light from water into air, object distance u' = -0.4 m): 1/v' - (4/3)/u' = 2.22, so 1/v' = 2.22 - (4/3)/0.4 = 2.22 - 3.33 = -1.11 and v' = -0.9 m.
6. The negative sign shows that the final image is virtual, on the same side as the object for this refraction, that is, inside the tank.
7. Hence, the final image is formed 0.9 m from the lens, on the tank side (to the right of the lens), as a virtual image.

Teacher's Note:
a) The power of the lens between air and water is the same for light travelling in either direction.
b) Take care to treat the image formed by the lens as a virtual object for the mirror, since it lies behind the mirror.

12. A thin plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis as shown in figure. The separation between object and image planes is 1.8 m. The magnification of the image, formed by one of the half lens, is 2. Find the focal length of the lens and separation between the two halves. Draw the ray diagram for image formation.

Answer:
1. Both half-lenses form the image of the object on the same screen, so the object distance for one half equals the image distance for the other (the displacement method).
2. Let one half form a magnified image with magnification 2: v = 2u. The object and image planes are 1.8 m apart, so u + v = 1.8 m, that is, 3u = 1.8, giving u = 0.6 m and v = 1.2 m.
3. Focal length: 1/f = 1/u + 1/v (magnitudes, real image) = (0.6 + 1.2)/(0.6 × 1.2), so f = 0.72/1.8 = 0.4 m.
4. The other half-lens must be at the conjugate position, where the object distance is 1.2 m and the image distance 0.6 m (magnification 1/2).
5. The separation between the two halves along the axis = 1.2 - 0.6 = 0.6 m.
6. Ray diagram: draw the object at O on the axis, the first half-lens 0.6 m from it with a ray through its optical centre and a ray parallel to the axis refracted through the focus (0.4 m from the lens). They meet on the screen at 1.8 m from O, giving an inverted image twice the size of the object. The second half-lens, 0.6 m beyond the first, has its object distance 1.2 m and forms an inverted image half the size on the same screen.
7. Hence, f = 0.4 m and the separation between the halves is 0.6 m.

Teacher's Note:
a) This is the same condition as in the displacement method for a lens: the two positions give magnifications m and 1/m.
b) The ray diagram is drawn by taking two rays from the tip of the object for each half lens.

13. A right angle prism (45º -90º-45º) of refractive index n has a plane of refractive index n₁(n₁ < n) cemented to its diagonal face. The assembly is in air. The ray is incident at AB as shown in figure.
(a) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle.
(b) Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated.

Answer:
1. Let the angle of incidence at AB be i and the angle of refraction r, so sin i = n sin r. The ray inside the prism is inclined upward at angle r to the normal of AB (horizontal), and the normal to the diagonal AC makes 45° with the horizontal.
2. So the angle of incidence of the ray on the diagonal face is 45° - r.
3. (a) The ray strikes the diagonal at the critical angle C, where sin C = n₁/n. So 45° - r = C, that is, r = 45° - C.
4. sin i = n sin(45° - C) = n[(1/√2)cos C - (1/√2) sin C] = (1/√2)[√(n² - n₁²) - n₁], because n cos C = √(n² - n₁²) and n sin C = n₁.
5. So i = sin⁻¹{(1/√2)[√(n² - n₁²) - n₁]}.
6. (b) The ray passes the diagonal face undeviated when it falls normally on it, so 45° - r = 0 and r = 45°.
7. sin i = 1.352 × sin 45° = 1.352 × 0.7071 = 0.956, so i = 73°.
8. Hence, i = sin⁻¹{(1/√2)(√(n² - n₁²) - n₁)} for part (a) and i = 73° for part (b).

Teacher's Note:
a) A ray at normal incidence on a boundary passes without deviation whatever the second medium is.
b) The critical angle exists on the diagonal face because the ray goes from the denser prism to the rarer cement, n₁ < n.

14. A ray of light travelling in air is incident at grazing angle (incident angle = 90º) on a long rectangular slab of a transparent medium of thickness t = 1.0 m as shown in figure. The point of incidence is the origin A(0, 0). The medium has a variable index of refraction n(y) given by \( n(y) = [k y^{3/2} + 1]^{1/2} \)
where k = 1.0 \( (\text{meter})^{-3/2} \). The refractive index of air is 1.0.
(a) Obtain a relation between the slope of the trajectory of the ray at a point B(x, y) in the medium and the incident angle at that point.
(b) Obtain an equation for the trajectory y(x) of the ray in the medium.
(c) Determine the coordinates (x₁, y₁) of the point P, where the ray intersects the upper surface of the slab-air boundary.
(d) Indicate the path of the ray subsequently.

Answer:
1. The index depends only on y, so the medium can be treated as thin layers parallel to the x-axis, and n sin i is constant along the ray, where i is the angle that the ray makes with the y-axis (the normal to the layers).
2. (a) At B the ray makes the angle i with the y-axis, so the slope of the trajectory is dy/dx = cot i.
3. (b) At A: n = 1 and i = 90°, so n(y) sin i = 1 × sin 90° = 1. Then sin i = 1/n and cot i = √(n² - 1) = √(k y^3/2) = y^3/4 for k = 1.
4. So dy/dx = y^3/4, that is, y^-3/4 dy = dx. Integrating, 4y^1/4 = x, since x = 0 when y = 0. Equivalently, y = (x/4)⁴.
5. (c) At the upper surface y = t = 1.0 m, so x = 4 × 1^1/4 = 4.0 m. P is at (4.0 m, 1.0 m).
6. (d) At P, n = √(1 + 1) = √2, so sin i = 1/√2 and i = 45°. For refraction into air, 1 × sin e = √2 sin 45° = 1, so e = 90°. The ray emerges parallel to the boundary at P, that is, at grazing emergence, and travels along the surface.
7. Hence, the slope is cot i, the path is 4y^1/4 = x, P = (4.0 m, 1.0 m) and the ray emerges grazing the upper surface.

Teacher's Note:
a) This is a ray in a graded-index medium, so the ray bends continuously towards the normal.
b) The condition n sin i = constant (Snell's law for layers) is the key to every part.

15. An image Y is formed of point object X by a lens whose optic axis is AB as shown in figure. Draw a ray diagram to locate the lens and its focus. If the image Y of the object X is formed by a concave mirror (having the same optic axis AB) instead of lens, draw another ray diagram to locate the mirror and its focus. Write down the steps of construction of the ray diagrams.

Answer:
1. Lens: Y and X are on opposite sides of the axis, so the image is real and inverted, and the lens is convex.
2. Step 1: Join X and Y by a straight line. The point where it cuts the axis AB is the optical centre O of the lens, because a ray through the optical centre goes straight.
3. Step 2: Draw the lens as a line through O, perpendicular to AB.
4. Step 3: Draw a ray from X parallel to the axis AB up to the lens. After refraction it must pass through Y. Join the point on the lens to Y; the point where this refracted ray cuts AB is the principal focus F of the lens.
5. Mirror: The real image Y is on the opposite side of the axis from X, so the mirror is concave.
6. Step 1: Draw Y', the mirror image of Y in the axis AB (the point at the same distance above the axis). Join X to Y' and extend it to meet the axis AB. This point is the pole P of the mirror, because the ray striking the pole and the reflected ray are equally inclined to the axis.
7. Step 2: Draw the mirror as a line through P, perpendicular to AB, with the reflecting side facing X and Y.
8. Step 3: Draw a ray from X parallel to the axis to the mirror; the reflected ray must pass through Y. The point where it cuts the axis is the principal focus F of the mirror.
9. The pole P of the mirror falls on the axis on the extension beyond A, to the left of the positions of X and Y; the lens centre O falls between A and B.

Teacher's Note:
a) In the lens figure O lies between the foot of X and the foot of Y, and the focus is on the same side as Y.
b) Both constructions use only the two facts: a ray through the optical centre or pole follows the symmetry rule, and a ray parallel to the axis passes through the focus.

16. A vessel ABCD of 10 cm width has two small slits s₁ and s₂ sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of s₁s₂. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure alongside. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled upto OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid.

Answer:
1. The source S is below the axis POQ, so the path from S to s₁ is longer than the path from S to s₂. Path difference = d sin θ ≈ d × (0.4/2.0) = 0.8 mm × 0.2 = 0.16 mm.
2. The identical glass plates add the same path to both beams, so they do not change the path difference.
3. For the central bright fringe the total path difference must be zero, so the point on CD must be nearer to s₁ by 0.16 mm. If y is its height above Q, then d y/D = 0.16 mm, so y = (0.16 × 10 cm)/0.8 = 2 cm.
4. The central bright fringe lies 2 cm above Q (on the side of s₁, opposite to the source).
5. With the liquid filled up to OQ (covering the lower slit s₂), the beam from s₂ travels 10 cm in the liquid and gets an extra optical path (μ - 1) × 10 cm.
6. The central fringe is at Q when this extra path balances the 0.16 mm: (μ - 1) × 100 mm = 0.16 mm, so μ - 1 = 0.0016.
7. μ = 1.0016.
8. Hence, the central bright fringe is 2 cm above Q, and the refractive index of the liquid is 1.0016.

Teacher's Note:
a) The central fringe always moves to the side opposite to the source, because the path difference from the source must be compensated.
b) Here the liquid raises the optical path of the lower beam, so it cancels the extra distance of s₁ from the source.

17. A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength λ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If λ = 648 nm, obtain the least value of t for which the rays interfere constructively.

Answer:
1. At the upper surface the light goes from air (1.0) to the layer (1.8), a denser medium, so the reflected ray suffers a phase change of π (an extra path of λ/2).
2. At the lower surface the light goes from the layer (1.8) to the glass (1.5), a rarer medium, so the reflection takes place without any phase change.
3. The path difference between the two reflected rays = 2 × 1.8 × t = 3.6 t (normal incidence), together with the λ/2 path from the phase change at the top.
4. For constructive interference, 3.6 t + λ/2 must be a whole number of wavelengths, so 3.6 t = (n - 1/2)λ, where n = 1, 2, 3, ...
5. The least thickness is for n = 1: 3.6 t = λ/2 = 324 nm, so t = 324/3.6 = 90 nm.
6. Hence, the condition is 3.6 t = (n - 1/2)λ and the least thickness is 90 nm.

Teacher's Note:
a) The two reflections are from different kinds of boundaries: the first has a phase change and the second has none.
b) Check: with t = 90 nm the optical thickness 2μt = 324 nm = λ/2.

18. The young's double slit experiment is done in a medium of refractive index 4/3. A light of 600 nm wavelength is falling on the slits having 0.45 mm separation. The lower slit S₂ is covered by a thin glass sheet of thickness 10.4 μm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown.
(a) Find the location of central maximum (bright fringe with zero path difference) on the y-axis.
(b) Find the light intensity at point O relative to the maximum fringe intensity.
(c) Now if 600 nm light is replaced by white light of range 400 to 700 nm. Find the wavelength of the light that form maxima exactly at point O.
[All wavelengths in the problem are for the given medium of refractive index 4/3. Ignore dispersion]

Answer:
1. The wavelength 600 nm is in the medium, so the relative index of the sheet with respect to the medium is 1.5/(4/3) = 9/8. The extra optical path (in the medium) introduced by the sheet is Δ = (9/8 - 1) × 10.4 μm = 1.3 μm.
2. (a) The central maximum is where the path difference is zero, so it shifts towards the covered slit S₂ (downwards) by y = Δ D/d = (1.3 × 10⁻⁶ × 1.5)/(0.45 × 10⁻³) = 4.33 × 10⁻³ m. So the central maximum is at y = -4.33 mm.
3. (b) At O the path difference is Δ = 1.3 μm = 2.1667 λ, so the phase difference is φ = 2π × (1/6) = π/3 (apart from whole cycles).
4. I = I_max cos²(φ/2) = I_max cos²(π/6) = (3/4) I_max.
5. (c) A maximum at O requires Δ = mλ, so λ = 1300/m nm. For the range 400 to 700 nm: m = 2 gives 650 nm and m = 3 gives 433.33 nm (m = 1 gives 1300 nm and m = 4 gives 325 nm, outside the range).
6. Hence, the central maximum is at y = -4.33 mm, the intensity at O is (3/4) I_max and the wavelengths are 650 nm and 433.33 nm (in the medium).

Teacher's Note:
a) The wavelength in the medium is used throughout, so the sheet effect is the relative index multiplied by thickness.
b) The central fringe shifts towards the slit that is covered by the glass sheet.

19. A coherent parallel beam of microwaves of wavelength, λ = 0.5 mm falls on a young's double slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 0.1 m from it as shown in the figure.
(a) If the incident beam falls normally on the double slit apparatus, find the y-coordinates of all the interference minima on the screen.
(b) If the incident beam makes an angle of 30º with the x-axis (as in the dotted arrow shown in figure), find the y-coordinates of the first minima on either side of the central maximum.

Answer:
1. The figure shows the screen distance as D = 1 m, and the printed answers also use D = 1 m, so D = 1 m is used here. (If D = 0.1 m is used, every y-value below becomes ten times smaller.)
2. (a) For normal incidence the minima are at d sin θ = (2m + 1)λ/2, so sin θ = (2m + 1)(0.5)/(2 × 1.0) = 0.25(2m + 1). The values are sin θ = 0.25 and 0.75 (the next value, 1.25, is not possible).
3. y = D tan θ. For sin θ = 0.25, tan θ = 0.2582 and y = ±0.26 m. For sin θ = 0.75, tan θ = 1.134 and y = ±1.13 m.
4. (b) For oblique incidence the path difference between the slits is d (sin θ - sin 30°) for a point at angle θ above the axis. The central maximum is at sin θ = 0.5, that is, θ = 30°.
5. For the first minima, d (sin θ - sin 30°) = ±λ/2 = ±0.25 mm, so sin θ = 0.5 ± 0.25, which gives 0.75 and 0.25.
6. The y-coordinates of the first minima are D tan θ = 1.13 m and 0.26 m (on either side of the central maximum, which is at y = 1 × tan 30° = 0.58 m).
7. Hence, (a) y = ±0.26 m and ±1.13 m, and (b) y = 0.26 m and 1.13 m.

Teacher's Note:
a) Tilting the beam moves the central maximum to y = D tan 30°, and the first minima on its two sides fall at sin θ = 0.25 and 0.75.
b) The wavelength is only 0.5 mm, so the maximum possible sin θ gives only two minima on each side.

20. In a Young's experiment, the upper slit is covered by a thin glass plate of refractive index 1.4, while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen, where the central maximum (n = 0) fall before the glass plates were inserted, now has 3/4 the original intensity. It is further observed that what used to be the fifth maximum lies below the point P while the sixth minima lies above P. Calculate the thickness of glass plate. (absorption of light by glass plate may be neglected).

Answer:
1. Let the thickness of each plate be t. The extra optical path of the lower beam relative to the upper = (1.7 - 1.4)t = 0.3 t = Δ. Because the lower beam is delayed, the whole pattern shifts downwards (towards the lower slit) by Δ/λ fringe widths.
2. At P the path difference is now Δ, so the intensity is I = I₀ cos²(φ/2) with φ = 2πΔ/λ. Since I = (3/4) I₀, cos²(φ/2) = 3/4, so φ/2 = π/6 (plus a multiple of π), that is, Δ/λ = N + 1/6 or N + 5/6, where N is a whole number.
3. The pattern has shifted down by s = Δ/λ fringes. The original fifth maximum (at 5 fringes) lies below P, so s is more than 5. The sixth minimum (at 5.5 fringes) lies above P, so s is less than 5.5.
4. Hence 5 < s < 5.5, and the only value of the form N + 1/6 in this range is s = 5 + 1/6 = 31/6.
5. Δ = (31/6)λ = (31/6) × 5400 × 10⁻¹⁰ m = 2.79 × 10⁻⁶ m.
6. t = Δ/0.3 = 2.79 × 10⁻⁶/0.3 = 9.3 × 10⁻⁶ m.
7. Hence, the thickness of each glass plate is 9.3 μm.

Teacher's Note:
a) The intensity 3/4 of the maximum fixes only the fractional part (1/6) of the shift; the whole number comes from the other two observations.
b) Check: 0.3 × 9.3 μm = 2.79 μm = 5.1667 × 0.54 μm.

21. In Young's experiment, the source is red light of wavelength 7 × 10^-7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by 10^-3 m to the position previously occupied by the 5th bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 × 10^-7 m, the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of glass for the green light. Also estimate the change in fringe width due to the change in wavelength.

Answer:
1. A plate of thickness t and index μ shifts the central fringe by (μ - 1) t, measured as a path difference, which equals a shift of (μ - 1)t/λ fringes.
2. For red light the shift is 5 fringes: (1.5 - 1) t = 5 × 7 × 10⁻⁷ m, so t = 3.5 × 10⁻⁶/0.5 = 7 × 10⁻⁶ m.
3. The red fringe width is β_r = (shift)/5 = 10⁻³/5 = 2 × 10⁻⁴ m.
4. For green light the central fringe moves to the place of the 6th red bright fringe, a shift of 6 β_r on the screen. The shift is proportional to (μ_g - 1)t, and the red shift (5β_r) corresponds to (μ - 1)t = 0.5 t. So (μ_g - 1) t = (6/5) × 0.5 t = 0.6 t, giving μ_g - 1 = 0.6 and μ_g = 1.6.
5. Fringe width is proportional to λ: β_g = β_r × (5/7) = 1.43 × 10⁻⁴ m.
6. Change in the fringe width = β_g - β_r = (5/7 - 1) × 2 × 10⁻⁴ = -5.71 × 10⁻⁵ m, that is, a decrease of 5.71 × 10⁻⁵ m.
7. Hence, the thickness is 7 × 10⁻⁶ m, the refractive index for green light is 1.6 and the fringe width decreases by 5.71 × 10⁻⁵ m.

Teacher's Note:
a) The refractive index is larger for the shorter wavelength, which is consistent with 1.6 for green and 1.5 for red.
b) The shift on the screen equals (μ - 1)tD/d, and the same D/d is used for both colours.

23. In figure, S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L₁ and L₂ by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 mm. The distance along the axis from S to L₁ and L₂ is 0.15 m while that from L₁ and L₂ to O is 1.30 m. The screen at O is normal to SO.
(a) If the third intensity maximum occurs at the point A on the screen, find the distance OA.
(b) If the gap between L₁ and L₂ is reduced from its original value of 0.5 mm, will the distance OA increase, decrease, or remain the same?

Answer:
1. Each half-lens forms its own real image of S, and these two images act as two coherent sources. Object distance u = 0.15 m and f = 0.10 m, so 1/v = 1/0.10 - 1/0.15 = 1/0.30 and v = 0.30 m. The magnification is 0.30/0.15 = 2 (inverted).
2. The optical centre of each half is 0.25 mm from the axis (half of the 0.5 mm gap), so S is 0.25 mm below the axis of L₁. The image from L₁ is therefore 2 × 0.25 = 0.5 mm above the axis of L₁, that is, at 0.25 + 0.5 = 0.75 mm above the axis SO. Similarly the image from L₂ is 0.75 mm below SO.
3. The two coherent sources are separated by d' = 1.5 mm, at a distance 0.30 m from the lens, so their distance from the screen is D' = 1.30 - 0.30 = 1.00 m.
4. (a) The third maximum is at y = 3λD'/d' = (3 × 500 × 10⁻⁹ × 1.00)/(1.5 × 10⁻³) = 1.0 × 10⁻³ m. So OA = 1 mm.
5. (b) If the gap g is reduced, the separation of the images d' = 3g also reduces (d' = 3 × 0.5 mm = 1.5 mm), while D' stays the same. Since OA = 3λD'/d' = λD'/g is inversely proportional to the gap, OA increases.
6. Hence, OA = 1 mm, and OA increases when the gap is reduced.

Teacher's Note:
a) This is the arrangement of Billet's split lens, in which the two halves give two real images that behave as coherent sources.
b) The distance of the sources from the screen is measured from the images, not from the lens.