Q: In a Young's double slit experiment, a combination of two glass wedges A and B, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separ

aligned & x+y=12 m \\ & 4 12 = 1 x \\ & x=3 m \\ & y=6 m aligned aligned & = y d D - ( _B-1 ) x- ( _A-1 ) y+ ( _B-1 ) y+ ( _A-1 ) x \\ & -y d D =-0.5 3-0.7 9+0.5 9+0.7 3 \\ & -y d D =-0.5 6-0.7 6 \\ & -y d D =-1.2 ~m \\ & y= 1.2 D d = 1.2 2 2 10^ -3 10^ -6 \\ & =1.2 ~mm aligned

Q: Paragraph I: In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 ~mm . The distance b

aligned & y=n ( D d ) \\ & for 8^ th fringe \\ & y=8 D d \\ & y_ =8 D d_ \\ & y_ =8 D d_ \\ & y_ -y_ =8 D [ 1 d_ - 1 d_ ] \\ & =6000 Ã… aligned aligned & D =1 ~m \\ & ~d _ =0.34 ~mm \\ & ~d _ min =0.76 ~mm \\ & y_ -y_ =8 6000 10^ -10 1 [ 1 0.76 10^ -3 - 1 0.84 10^ -3 ] \\ & =8 6 10^ -4 [ 0.08 0.76 0.84 ]=601.5 m \\ & aligned (answer range is from 598 to 602 )

Q: In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 ~mm . The distance between the sl

aligned & y = n D d \\ & v = dy dt =- n d d ^2 d(d) d t \\ & d=0.8+0.04 t \\ & d(d) d t =0.04 t aligned for v d ( d ) dt . For d ( d ) dt . aligned & t =1 t =0 \\ & ( d ( d ) dt )_ =0.04 \\ & d =0.8 ~mm \\ & v _ = 8 6000 10^ -10 1 0.04 0.08 0.8 0.8 10^ -6 10^ -3 =24 m / s . aligned (range is from 23.5 to 24.5)

Q: A point source S emits unpolarized light uniformly in all directions. At two points A and B , the ratio r=I_A / I_B of the intensities of light is 2. If a set of two polaroids having 45^ angle between

$I 1 l^2 $ Where I: distance from point source. $ aligned & I_A I_B = I_B^2 I_A^2 =2 \\ & I_B= 2 I_A . aligned $ Also, due to polaroids: $ aligned & I_B^ = I_B 2 ^2 45^ = I_B 4 \\ & I_B^ = I_B 4 (2) \\ & Ratio becomes 4 times. \\ & r_ new =8 aligned $

Q: A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n 2 . The other slit is at the interface of this medium with another medium 1 of refractive

From the above figure, A B = d tan θ and B C = A B sin α = d tan θ sin α Also, A D = A B sin θ ⇒ Path difference (in vaccum) = n 1 B C - n 2 A D = n 1 A B sin α - n 2 A B sin θ From snell's law n 1 sin α = n 2 sin θ . Therefore, path difference = A B n 1 sin α - n 2 sin θ = 0 ⇒ A , B are correct C , D are incorrect

Question 1

In the List-I, four optical effects are mentioned. The physical phenomena of light which are essential to describe these optical effects are given in List-II. Choose the option which describes the cor

Accepted Answer

The colorful sky in the north polar region, known as Aurora Borealis, is caused by the emission of radiation from oxygen and nitrogen atoms in the upper atmosphere when they are excited by charged particles from the solar wind. Thus, P 5. Sunlight scattered by molecules in the atmosphere becomes partially polarized in a direction perpendicular to the incident light. Thus, Q 4. A rainbow is formed due to the dispersion of sunlight and its internal reflection by water droplets in the atmosphere. T

Question 2

In a single slit diffraction experiment, a slit of width (0.016 0.002) mm is used to measure the wavelength of a monochromatic light source. In the diffraction pattern, the angular distance between th

Accepted Answer

For a single slit diffraction, the condition for the first minimum is given by: a = Taking the natural logarithm on both sides, we get: = a + ( ) Differentiating to find the maximum fractional error: = a a + Since = 2^ is very small, 1, which gives 1 . Given (2^ ) = 0.035, we can approximate the angle in radians as = 0.035 rad. The error in the angle is = 40' = ( 40 60 )^ = ( 2 3 )^ . Converting into radians: = 2 3 ( 0.035 2 ) = 0.035 3 rad The fractional error in the angular term is: = 0.035 3

Question 3

As shown in the figure, a ray AB of unpolarized light enters from water of refractive index n_w = 4/3 into a medium of refractive index n_p = 4/ 3 after passing through a glass plate of refractive ind

Accepted Answer

Let the angle of incidence at the first interface (water to glass) be i. By applying Snell's law at the successive parallel interfaces, we can find the angle of incidence at the interface where the reflection occurs (water to medium n_p). For the first interface (water to glass): n_w i = n_g r_1 For the second interface (glass to water): n_g r_1 = n_w r_2 From the above two equations, we get: n_w i = n_w r_2 r_2 = i Thus, the angle of incidence at the third interface (water to medium n_p) is als

Question 4

A single slit diffraction experiment is performed to determine the slit width using the equation, b d D =m , where b is the slit width, D the shortest distance between the slit and the screen, d the d

Accepted Answer

Original question asked about absolute error which can be both minimum or maximum, we added word maximum to make it clear for students aligned & b = m D d =360 ~m \ & ~b _ = 3 600 10^ -3 1.01 4 10^ -3 ~m =454.5 ~m \ & ~b _ = 3 600 10^ -3 0.99 6 10^ -3 ~m =297 ~m aligned Maximum value of b gives error, b _1=94.5 ~m Minimum value of b gives error, b_2=63 ~m We always report the largest error, hence correct answer should be 94.5 ~m

Question 5

In a Young's double slit experiment, a combination of two glass wedges A and B, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separ

Accepted Answer

aligned & x+y=12 m \ & 4 12 = 1 x \ & x=3 m \ & y=6 m aligned aligned & = y d D - ( _B-1 ) x- ( _A-1 ) y+ ( _B-1 ) y+ ( _A-1 ) x \ & -y d D =-0.5 3-0.7 9+0.5 9+0.7 3 \ & -y d D =-0.5 6-0.7 6 \ & -y d D =-1.2 ~m \ & y= 1.2 D d = 1.2 2 2 10^ -3 10^ -6 \ & =1.2 ~mm aligned

Question 6

Paragraph I: In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 ~mm . The distance b

Accepted Answer

aligned & y=n ( D d ) \ & for 8^ th fringe \ & y=8 D d \ & y_ =8 D d_ \ & y_ =8 D d_ \ & y_ -y_ =8 D [ 1 d_ - 1 d_ ] \ & =6000 Ã… aligned aligned & D =1 ~m \ & ~d _ =0.34 ~mm \ & ~d _ min =0.76 ~mm \ & y_ -y_ =8 6000 10^ -10 1 [ 1 0.76 10^ -3 - 1 0.84 10^ -3 ] \ & =8 6 10^ -4 [ 0.08 0.76 0.84 ]=601.5 m \ & aligned (answer range is from 598 to 602 )

Question 7

In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 ~mm . The distance between the sl

Accepted Answer

aligned & y = n D d \ & v = dy dt =- n d d ^2 d(d) d t \ & d=0.8+0.04 t \ & d(d) d t =0.04 t aligned for v d ( d ) dt . For d ( d ) dt . aligned & t =1 t =0 \ & ( d ( d ) dt )_ =0.04 \ & d =0.8 ~mm \ & v _ = 8 6000 10^ -10 1 0.04 0.08 0.8 0.8 10^ -6 10^ -3 =24 m / s . aligned (range is from 23.5 to 24.5)

Question 8

A point source S emits unpolarized light uniformly in all directions. At two points A and B , the ratio r=I_A / I_B of the intensities of light is 2. If a set of two polaroids having 45^ angle between

Accepted Answer

$I 1 l^2 $ Where I: distance from point source. $ aligned & I_A I_B = I_B^2 I_A^2 =2 \ & I_B= 2 I_A . aligned $ Also, due to polaroids: $ aligned & I_B^ = I_B 2 ^2 45^ = I_B 4 \ & I_B^ = I_B 4 (2) \ & Ratio becomes 4 times. \ & r_ new =8 aligned $

Question 9

A double slit setup is shown in the figure. One of the slits is in medium  2 of refractive index n 2 . The other slit is at the interface of this medium with another medium  1 of refractive

Accepted Answer

From the above figure, A B = d tan &#952; and  B C = A B sin &#945; = d tan &#952; sin &#945; Also,  A D = A B sin &#952; &#8658;  Path difference (in vaccum) = n 1 B C - n 2 A D = n 1 A B sin &#945; - n 2 A B sin &#952; From snell's law  n 1 sin &#945; = n 2 sin &#952; . Therefore, path difference = A B n 1 sin &#945; - n 2 sin &#952; = 0 &#8658; &#160; &#160; A , B  are correct C , D  are incorrect

Question 10

A parallel beam of light strikes a piece of transparent glass having cross section as shown in the figure below. Correct shape of the emergent wavefront will be (figures are schematic and not drawn to

Accepted Answer

As we can see that the thickness of the transparent glass is minimum for the rays which are in the middle portion. So these will have the minimum optical path and the rays on the extreme ends will have to cover the maximum optical path length. Hence, the wavefront of wave 2 will be ahead of waves 1 and 3.

Wave Optics — Previous Year Questions & Solutions