Q: Two identical concave mirrors each of focal length f are facing each other as shown in the schematic diagram. The focal length f is much larger than the size of the mirrors. A glass slab of thickness

aligned & S-t 2 + t 2 n_0 =2 f \\ & S+t ( 1 n_0 -1 )=4 f \\ & S=4 f+ (1- 1 n_0 ) t aligned Also aligned & S - t 2 + t 2 n _0 = f \\ & S =2 f + (1- 1 n _0 ) t aligned

Q: Two equilateral-triangular prisms P _1 and P _2 are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism P_1 at an angle of incidence such that the

At surface BC aligned & 3 2 r _2= 3 30 \\ & 3 2 r _2= 3 2 \\ & sinr _2= 1 2 \\ & r _2=45^ \\ & r _1=60^ -45^ =15^ aligned At surface A B array ll 1 = 3 2 15^ \\ = ^ -1 [ 3 2 12 ] \\ =12 array

Q: A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence _0. The ray partially refracts into the sphere with angle of refraction _0 and then partly reflects fr

aligned & = ( _0- _0 )+ (180-2 _0 )+ ( _0-2 _0 ) \\ & =180+2 _0-4 _0 aligned (P) =180+2 _0-4 _0 180=180+2 _0-4 _0 _0=2 _0 ...(i) _0=2 _0 ...(ii) From (i) & (ii) aligned & _0=2 ( _0 / 2 ) ( _0 2 )=1 \\ & _0 2 =0 \\ & _0=0 aligned (Q) _0=2 _0 ...(i) _0= 3 _0 ...(ii) From (i) & (ii) aligned & _0= 3 ( _0 2 ) \\ & ( _0 2 )= 3 2 \\ & _0 2 =30,150 \\ & _0=60,300 (Rejected) \\ & _0=60 , 0 aligned (R) _0=2 _0 aligned & _0= 3 _0 \\ & 2 _0= 3 _0 \\ & _0= 3 2 \\ & _0=30,150 (Rejected) aligned _0=30, 0 ...(i

Q: A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is δ = 60 o (see Figure- 1 ). Th

For no reflection: tan i = 3 ⇒ i = 60 ° Applying Snell's law at left interface, we get ⇒ 1 × sin i = 3 × sin r 1 ⇒ sin r 1 = 1 2 ⇒ r 1 = 30 o Now, for given values δ = i + e - A ⇒ 60 ° = 60 ° + e - A ⇒ e = A Then, A = r 1 + r 2 ⇒ r 2 = A - r 1 = 30 o For red light: Expression to find minimum deviation is given by, μ = sin A + δ m 2 sin A 2 = sin 60 ° + 30 ° 2 sin 60 ° 2 = 1 2 1 2 = 2

Question 1

A double convex lens made of glass of refractive index 1.5 and radii of curvature of the curved surfaces 20 cm each is immersed in a liquid of refractive index n_L. The correct plot showing the variat

Accepted Answer

The focal length f of a lens immersed in a medium of refractive index n_L is given by the Lens Maker's formula: 1 f = ( n_g n_L - 1 ) ( 1 R_1 - 1 R_2 ) Given that the lens is double convex, the radii of curvature are R_1 = +20 cm = +0.2 m and R_2 = -20 cm = -0.2 m . The refractive index of the glass is n_g = 1.5. Substituting these values into the formula gives: 1 f = ( 1.5 n_L - 1 ) ( 1 0.2 - 1 -0.2 ) 1 f = ( 1.5 n_L - 1 ) (5 + 5) = 10 ( 1.5 n_L - 1 ) The power P of the lens is defined as the r

Question 2

A beam of polychromatic light passes through a thin prism of prism angle 6^ . The refractive index of the material of the prism varies with wavelength ( ) as n( ) = + ^2 , where = 3\ m ^ -1 and = 0.09

Accepted Answer

For a thin prism, the angle of deviation is given by D = (n - 1)A. To find the smallest value of deviation D_m, we need to find the minimum value of the refractive index n( ). Given n( ) = + ^2 . Differentiating n( ) with respect to and equating to zero for minimum: dn d = - 2 ^3 = 0 ^3 = 2 Substituting the given values = 3\ m ^ -1 and = 0.096\ m ^2: ^3 = 2 0.096 3 = 0.064 = 0.4\ m Now, substituting = 0.4\ m back into the expression for n( ): n_ = 3(0.4) + 0.096 (0.4)^2 n_ = 1.2 + 0.096 0.16 = 1

Question 3

Consider two isosceles prisms 1 and 2 with prism angles A_1 and A_2 and refractive indices n_1 and n_2, respectively, as shown in the figure. The faces a_1 b_1 and a_2 b_2 are parallel to each other a

Accepted Answer

From the given geometry, faces a_1b_1 and a_2b_2 are parallel to each other and perpendicular to the horizontal mirror M. Thus, a_1b_1 and a_2b_2 are vertical. The normal to a_1b_1 is horizontal. The ray emerging from Prism 1 makes an angle e_1 with this horizontal normal. By the law of reflection at mirror M, the reflected ray also makes an angle e_1 with the horizontal. This reflected ray is incident on the vertical face a_2b_2 of Prism 2. The normal to a_2b_2 is horizontal, so the angle of in

Question 4

A solid glass sphere of refractive index n= 3 and radius R contains a spherical air cavity of radius R 2 , as shown in the figure. A very thin glass layer is present at the point O so that the air cav

Accepted Answer

We added a condition θ > 30° to make this question clear for students, otherwise 0.5 and 0.75 both are correct answers aligned & = 3 \ & =60^ aligned aligned & 3 =1 =30^ \ & R 2 30^ = x 120^ \ & R 120^ = R 3 2 = 3 2 3 2 \ & = 3 4 aligned

Question 5

Two identical concave mirrors each of focal length f are facing each other as shown in the schematic diagram. The focal length f is much larger than the size of the mirrors. A glass slab of thickness

Accepted Answer

aligned & S-t 2 + t 2 n_0 =2 f \ & S+t ( 1 n_0 -1 )=4 f \ & S=4 f+ (1- 1 n_0 ) t aligned Also aligned & S - t 2 + t 2 n _0 = f \ & S =2 f + (1- 1 n _0 ) t aligned

Question 6

Two equilateral-triangular prisms P _1 and P _2 are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism P_1 at an angle of incidence such that the

Accepted Answer

At surface BC aligned & 3 2 r _2= 3 30 \ & 3 2 r _2= 3 2 \ & sinr _2= 1 2 \ & r _2=45^ \ & r _1=60^ -45^ =15^ aligned At surface A B array ll 1 = 3 2 15^ \ = ^ -1 [ 3 2 12 ] \ =12 array

Question 7

A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The radius of curvature of the convex surface (SPU) is 9 ~cm , while the planar surface (STU) acts as a

Accepted Answer

Since STU is a plane mirror, we can take mirror image of the whole situation about it and final image can be assumed to be at a distance h below the base. Since object and image are at same distance from equivalent lens, hence h =2 ~F _ eq aligned & 1 ~F _ eq = ( 1.6-1 1 ) ( 2 9 )+ ( n -1) 1 ( -2 9 ) \ & 1 h 2 = 1.2 9 + 2(1- n ) 9 \ & 2 ~h = 3.2-2 n 9 \ & ~h = 9 1.6- n cm aligned (1) for n =1.42, ~h =50 ~cm (2) for n =1.35, ~h =36 ~cm (3) for n =1.45, ~h =60 ~cm (4) for n =1.48, ~h =75 ~cm

Question 8

A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence _0. The ray partially refracts into the sphere with angle of refraction _0 and then partly reflects fr

Accepted Answer

aligned & = ( _0- _0 )+ (180-2 _0 )+ ( _0-2 _0 ) \ & =180+2 _0-4 _0 aligned (P) =180+2 _0-4 _0 180=180+2 _0-4 _0 _0=2 _0 ...(i) _0=2 _0 ...(ii) From (i) & (ii) aligned & _0=2 ( _0 / 2 ) ( _0 2 )=1 \ & _0 2 =0 \ & _0=0 aligned (Q) _0=2 _0 ...(i) _0= 3 _0 ...(ii) From (i) & (ii) aligned & _0= 3 ( _0 2 ) \ & ( _0 2 )= 3 2 \ & _0 2 =30,150 \ & _0=60,300 (Rejected) \ & _0=60 , 0 aligned (R) _0=2 _0 aligned & _0= 3 _0 \ & 2 _0= 3 _0 \ & _0= 3 2 \ & _0=30,150 (Rejected) aligned _0=30, 0 ...(i

Question 9

A monochromatic light wave is incident normally on a glass slab of thickness d , as shown in the figure. The refractive index of the slab increases linearly from n 1  to n 2 over the height h . W

Accepted Answer

Dashed line M N and P Q are incident wave front. Just before entering the slab P Q is maxima. P 1 and Q 1 are the points on other face of slab. So, P 1 will be at a lead of n 2 d w.r.t. P and Q '  will be at lead of n 1 d w.r.t. Q As the medium on the right side is homogeneous, so to have same path lead as P 1 , &#160; Q '  must travel to Q 1 , such that n 2 d = n 1 d + h &#160; sin &#952; . sin &#952; = n 2 - n 1 d h &#8658; &#952; = sin - 1 n 2 - n 1 d h

Question 10

A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is &#948; = 60 o (see Figure- 1 ). Th

Accepted Answer

For no reflection: tan i = 3 &#8658; i = 60 &#176; Applying Snell's law at left interface, we get &#8658; 1 &#215; sin i = 3 &#215; sin r 1 &#8658; sin r 1 = 1 2 &#8658; r 1 = 30 o Now, for given values &#948; = i + e - A &#8658; 60 &#176; = 60 &#176; + e - A &#8658; e = A Then,  A = r 1 + r 2 &#8658; r 2 = A - r 1 = 30 o For red light: Expression to find minimum deviation is given by, &#956; = sin A + &#948; m 2 sin A 2 = sin 60 &#176; + 30 &#176; 2 sin 60 &#176; 2 = 1 2 1 2 = 2

Question 11

An optical arrangement consists of two concave mirrors M 1  and M 2 , and a convex lens L with a common principal axis, as shown in the figure. The focal length of L is 10 &#160; cm . The radii o

Accepted Answer

For reflection from M 2 : Applying mirror formula, we get 1 v + 1 - 10 = 1 - 12 &#8658; 1 v = 1 10 - 1 12 &#8658; v = + 60 &#160; cm  (for I 1 ) For refraction from L : Applying lens equation, we get 1 v - 1 u = 1 f &#8658; 1 v - 1 - 80 = 1 10 &#8658; v = + 80 7  (for  I 2 ) This image should be at focus of M 1 &#8756; &#160; 20 2 + 80 7 = n 7 &#8658; n = 150 Also, If I 2  is formed at pole of M 1 then  n 7 = 80 7 &#8658; n = 80 And further if I 2  is formed at cent

Question 12

Consider a configuration of n  identical units, each consisting of three layers. The first layer is a column of air of height h = 1 3 &#160; cm , and the second and third layers are of equal thic

Accepted Answer

Considering a single unit with three layers. The figure below shows the refraction of light as it moves from one medium to others. For the first interface,  x 1 = 1 3 &#215; tan 60 &#176; = 1 3 &#160; cm Using Snell's law, &#956; 1 sin &#952; 1 = &#956; 2 sin &#952; 2 &#8658; 1 &#215; 3 2 = 3 2 &#215; sin &#952; 2 &#8658; &#952; 2 = 45 &#176; For the second interface, &#8658; x 2 = d &#215; tan 45 &#176; = d And again using Snell's law, 3 2 &#215; sin 45 = 3 &#215; sin &#952; 3 &#86

Question 13

An object and a concave mirror of focal length f = 10 &#160; cm  both move along the principal axis of the mirror with constant speeds. The object moves with speed V 0 = 15 &#160; cm &#160; s - 1

Accepted Answer

Magnification expression for concave mirror is given by, m = f f - u = - 10 - 10 - - 30 = - 1 2 For concave mirror, we can write, v i , m = - m 2 v o . m &#8658; &#160; v o - v m &#215; m 2 + v i - v m = 0 As given in the question,  v i = 0 . Therefore,  &#8658; &#160; v o - v m &#215; 1 4 = v m &#8658; v o = 5 v m &#8658; v m = v o 5 = 15 5 = 3 &#160; cm &#160; s - 1

Question 14

A rod of length 2   cm makes an angle 2 &pi; 3 rad with the principal axis of a thin convex lens. The lens has a focal length of 10   cm and is placed at a distance of 40 3   cm from th

Accepted Answer

In convex lens, when an object is placed in between the centre of curvature and focus, the real image is formed behind the centre of curvature. The size of the image is larger than that of the object. Here, the angle made by inner side of rod with principal axis is  &#960; - 2 &#960; 3 = &#960; 3 &#160; rad . From figure, height of object is  h o = 2 sin 60 &#176; = 2 &#215; 3 2 = 3 &#160; cm  and given height of image  h i = 30 3 13 &#160; cm Size of object is  u =

Question 15

Three plane mirrors form an equilateral triangle with each side of length L . There is a small hole at a distance l > 0  from one of the corners as shown in the figure. A ray of light is passe

Accepted Answer

As we can see, for &#952; = 30 &#176; , the ray will incident normally and hence will retrace its path, hence the ray will come out for  &#952; = 30 &#176;  and  0 &#60; l &#60; L . &#8658; A is correct Considering the symmetry of the situation, we can have: Thus, for  &#952; = 60 &#176;  and  l = L 2 , ray will come out from the setup after two reflections. &#8658; B  is correct Now, when  &#952; = 60 &#176;  and  l = L 3 , the ray on enter

Question 16

List I contains four combinations of two lenses ( 1 and 2 ) whose focal lengths (in cm ) are indicated in the figures. In all cases, the object is placed 20 cm from the first lens on the left, and the

Accepted Answer

I u = &#8211; 20 &#160; cm   f 1 = + 10 &#160; cm Using the lens formula, 1 v - 1 u = 1 f &#8658; 1 v 1 + 1 20 = 1 10 &#8658; 1 v 1 = 1 10 - 1 20 &#8658; v 1 = 20 &#160; cm For second lens  u = v 1 - 5 = + 15 &#160; cm ,  f 2 = + 15 &#160; cm &#8658; 1 v - 1 u = 1 f 2 &#8658; 1 v - 1 15 = 1 15 &#8658; 1 v = 2 15 v = + 7 . 5 &#160; cm (from lens 2 ) Therefore,  I &#8594; P (II)  Similarly, u = - 20 &#160; cm , f = + 10 &#160; cm 1 v + 1 20 = 1 10 &#8658; 1 v = 1 10 - 1 20

Question 17

For a prism of prism angle  &#952; = 60 &#176; , the refractive indices of the left half and the right half are, respectively,  n 1  and  n 2 ,  n 2 &#8805; n 1  as shown

Accepted Answer

Given,  n 1 = n 2 = n = 3 2 In case of minimum deviation, r m = A 2 = 60 &#176; 2 = 30 &#176; and  i = e Applying expression for minimum deviation, n = sin &#948; min + A 2 sin A 2 &#8658; 3 2 = sin ( i ) sin 60 &#176; 2 &#8658; sin i = 3 4 , &#160;&#160; cos i = 7 4 If  n 1 = n = 3 2  and  n 2 = n + &#8710; n : Angle of emergence,  e = i + &#916; e Applying Snell's law, we get, 1 &#215; sin i = n sin 30 &#176; Then, n + &#916; n sin 30 &#176; = 1 sin i + &#916;

Question 18

An extended object is placed at point  O ,  10 &#160; cm  in front of a convex lens  L 1  and a concave lens  L 2  is placed  10 &#160; cm  behind it, as s

Accepted Answer

For convex lens, 1 f = &#956; 2 &#956; 1 &#8722; 1 1 R 1 &#8722; 1 R 2 &#8658; 1 f 1 = 3 2 &#8722; 1 1 20 &#8722; 1 - 20 &#8658; f 1 = + 20 &#160; cm Now, image distance from convex, v convex = u f 1 u + f 1 = &#8722; 10 &#215; 20 &#8722; 10 + 20 &#8658; v convex = &#8722; 20 &#160; cm This location of image will be object location for concave lens. Therefore,  u concave = &#8722; 30 &#160; cm For concave lens, 1 f 2 = 3 2 &#8722; 1 1 &#8722; 20 &#8722; 1 20 &#8658; f 2 = &#8722; 20 &#160;

Question 19

A wide slab consisting of two media of refractive indices  n 1  and  n 2  is placed in air as shown in the figure. A ray of light is incident from medium  n 1  to  n

Accepted Answer

Option A: If  n 1 = n 2 , n 1 sin &#952; = n 2 sin &#952; 1 &#160; &#160; &#160; &#160; &#160; &#160; . . . 1 and  n 2 sin &#952; 1 = 1 &#215; sin r &#160; &#160; &#160; &#160; &#160; . . . 2 From (1) and (2), sin r = n 1 sin &#952; Now,  sin &#952; &#62; 1 n 1 , (given) &#8756; &#160; sin r &#62; 1  (which is not possible) and hence light will reflect back from  air , &#160; n 2  interface Option B:  If  n 2 &#60; n 1 , Snell's law at  A , n

Question 20

A large square container with thin transparent vertical walls and filled with water refractive index 4 3 is kept on a horizontal table. A student holds a thin straight wire vertically inside the water

Accepted Answer

P 1 O = 12 2 cm P 1 I 1 = 12 2 × 4 3 = 9 2 cm O I 1 = 12 2 - 9 2 = 3 2 ; O I 2 = 3 2 cm I 1 I 2 = 3 2 × 1 2 × 2 = 3 The above solution is assumed according to the normal incidence. If we consider the actual angle of incidence, i.e., 45 ° the answer comes to be 3 .

Ray Optics — Previous Year Questions & Solutions