Wave Optics — Page 2 | JEE Physics

Q11

A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is

A circle

B hyperbola

C parabola

D straight line

Correct Answer

Option D

Solution

The interference fringes formed in a Young's double slit experiment are straight lines.

These fringes are a result of the constructive and destructive interference of the light waves coming from the two slits, and they appear as alternating bright and dark straight bands or lines.

Hence, the answer is : Option D : straight line.

Remember for double hole experiment a hyperbola is generated.

Q12

If

{I_0}

is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?

A

4{I_0}

B

2{I_0}

C

{{{I_0}} \over 2}

D

{I_0}

Correct Answer

Option D

Solution

I = {I_0}{\left( {{{\sin \theta } \over \theta }} \right)^2}

and

\theta = {\pi \over \lambda }\left( {{{ay} \over D}} \right)

For principal maximum y = 0 $\therefore$ $\theta$ = 0 When $\theta = 0$ , the intensity formula becomes : $I = I_0\left(\dfrac{\sin \theta}{\theta}\right)^2 = I_0\left(\dfrac{\sin 0}{0}\right)^2$ Here, we have an indeterminate form $\dfrac{0}{0}$ .

However, by taking the limit of the expression as $\theta$ approaches 0, we find that : $\lim\limits_{\theta \to 0} \left(\dfrac{\sin \theta}{\theta}\right) = 1$ Therefore, the intensity at the principal maximum remains the same : $I = I_0\left(\dfrac{\sin \theta}{\theta}\right)^2 = I_0(1)^2 = I_0$ Hence intensity will remain same.

Q13

In a Young's double slit experiment the intensity at a point where the path difference is

{\lambda \over 6}

(

\lambda

being the wavelength of light used ) is

I

. If

{I_0}

denotes the maximum intensity,

{I \over {{I_0}}}

is equal to

A

{3 \over 4}

B

{1 \over {\sqrt 2 }}

C

{{\sqrt 3 } \over 2}

D

{1 \over 2}

Correct Answer

Option A

Solution

The intensity of light at any point of the screen where the phase difference due to light coming from the two slits is $\phi$ is given by

I = {I_0}{\cos ^2}\left( {{\phi \over 2}} \right)\,\,

where

{I_0}

is the maximum intensity. NOTE : This formula is applicable when

{I_1} = {I_2}.

Here

\phi = {{\pi}/{3}}

$\therefore$

{I \over {{I_0}}} = {\cos ^2}{\pi \over 6} = {\left( {{{\sqrt 3 } \over 2}} \right)^2} = {3 \over 4}

Q14

This question has a paragraph followed by two statements, Statement

-1

and Statement

-2

. Of the given four alternatives after the statements, choose the one that describes the statements. A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plane. With monochromatic light, this film gives an interference pattern due to light, reflected from the top (convex) surface and the bottom (glass plate) surface of the film. Statement -

1

: When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of

\pi .

Statement -

2

: The center of the interference pattern is dark.

A Statement -

1

is true, Statement -

2

is true, Statement -

2

is the correct explanation of Statement -

1

B Statement -

1

is true, Statement -

2

is true, Statement -

2

is not the correct explanation of Statement -

1

C Statement -

1

is false, Statement -

2

is true

D Statement -

1

is true, Statement -

2

is false.

Correct Answer

Option B

Solution

Statement - 1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of π.

This statement is true.

According to the physics of wave optics, when light reflects from a denser medium (glass in this case) to a rarer medium (air), there is a phase change of π (or 180 degrees).

Statement - 2 : The center of the interference pattern is dark.

This statement is also true.

When a thin air film is formed between a convex lens and a plane surface, the path difference at the center of the pattern is zero.

However, due to the phase change of π at the lower surface (glass surface), the two interfering waves are out of phase at the center, resulting in destructive interference and creating a dark spot at the center.

So both statements are true.

However, statement - 2 is not a direct explanation of statement - 1.

Statement - 2 arises because of the combination of the path difference and the phase change mentioned in statement - 1, but it doesn't directly explain why the phase change occurs.

Hence, the answer is : Option B : Statement - 1 is true, Statement - 2 is true, Statement - 2 is not the correct explanation of Statement - 1.

Q15

In Young's double slit experiment , one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If

{{\rm I}_m}

be the maximum intensity, the resultant intensity

{\rm I}

when they interfere at phase difference

\phi

is given by :

A

{{{I_m}} \over 9}\left( {4 + 5\cos \,\phi } \right)

B

{{{I_m}} \over 3}\left( {1 + 2{{\cos }^2}\,{\phi \over 2}} \right)

C

{{{I_m}} \over 3}\left( {1 + 4{{\cos }^2}\,{\phi \over 2}} \right)

D

{{{I_m}} \over 9}\left( {1 + 8{{\cos }^2}\,{\phi \over 2}} \right)

Correct Answer

Option D

Solution

Let

{a_1} = a,\,{I_1} = a_1^2 = {a^2}

{a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}

Therefore

{{\rm I}_2} = 4{{\rm I}_1}

{I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi }

{I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi

\Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

Now,

{I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}

{I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}

Substituting in equation

\left( 1 \right)

{I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi

{I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]

{I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]

{I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]

Q16

Two beams,

A

and

B

, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam

A

has maximum intensity (and beam

B

has zero intensity), a rotation of polaroid through

{30^ \circ }

makes the two beams appear equally bright. If the initial intensities of the two beams are

{{\rm I}_A}

and

{{\rm I}_B}

respectively, then

{{{{\rm I}_A}} \over {{{\rm I}_B}}}

equals:

A

3

B

{3 \over 2}

C

1

D

{1 \over 3}

Correct Answer

Option D

Solution

According to malus law, intensity of emerging beam is given by,

I = {I_0}{\cos ^2}\theta

Now,

{I_{A'}} = {I_A}{\cos ^2}{30^ \circ }

{I_{B'}} = {I_B}{\cos ^2}{60^ \circ }

As

{I_{A'}} = {I_{B'}}

\Rightarrow {I_A} \times {3 \over 4} = {I_B} \times {1 \over 4}

$\therefore$

{{{I_A}} \over {{I_B}}} = {1 \over 3}

Q17

Assuming human pupil to have a radius of

0.25

cm

and a comfortable viewing distance of

25

cm

, the minimum separation between two objects that human eye can resolve at

500

nm

wavelength is :

A

100\,\mu m

B

300\,\mu m

C

1\,\mu m

D

30\,\mu m

Correct Answer

Option D

Solution

\sin \theta = {{0.25} \over {25}} = {1 \over {100}}

Resolving power

= {{1.22\lambda } \over {2\mu \sin \theta }} = 30\,\mu m.

Q18

Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is

\theta

. Then :

A cos

\theta

=

{\left( {{2 \over 3}} \right)^{{{1}/{2}}}}

B cos

\theta

=

{\left( {{2 \over 3}} \right)^{{{1}/{4}}}}

C cos

\theta

=

{\left( {{1 \over 3}} \right)^{{{1}/{2}}}}

D cos

\theta

=

{\left( {{1 \over 3}} \right)^{{{1}/{4}}}}

Correct Answer

Option B

Solution

As intensity of emergent beam is reduced to half after passing through two polarisers.

It means angle between A and B is 0

^\circ

. Now, on placing polariser C between A and B. Intensity after passing through A is

{I_A} = {I \over 2}

. Let $\theta$ be the angle between A and C. Intensity of light after passing through C is given by

{I_C} = {I \over 2}{\cos ^2}\theta

Intensity of light after passing through polariser B is

{I \over 3}

. Angle between C and B is also $\theta$ as A is parallel to B. So,

{I \over 3} = {I_C}{\cos ^2}\theta = {I \over 2}{\cos ^2}\theta \,.\,{\cos ^2}\theta

{\cos ^4}\theta = {2 \over 3} \Rightarrow \cos \theta = {\left( {{2 \over 3}} \right)^{1/4}}

Q19

Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star :-

A 610 × 10–9 radian

B 457.5 × 10–9 radian

C 305 × 10–9 radian

D 152.5 × 10–9 radian

Correct Answer

Option C

Solution

Limit of resolution of telescope

(\theta) = {{1.22\lambda } \over D}

\theta = {{1.22 \times 500 \times {{10}^{ - 9}}} \over {200 \times {{10}^{ - 2}}}} = {{1.22 \times 500 \times {{10}^{ - 9}}} \over 2}

$\theta$ = 305 × 10–9 radian

Q20

In a double-slit experiment, green light (5303

\mathop A\limits^ \circ

) falls on a double slit having a separation of 19.44

\mu

m and awidht of 4.05

\mu

m. The number of bright fringes between the first and the second diffraction minima is :

A 04

B 05

C 10

D 09

Correct Answer

Option B

Solution

For diffraction location of 1st minima y1 =

{{D\lambda } \over a}

= 0.2469 D $\lambda$ location of 2nd minima y2 =

{{2D\lambda } \over a}

= 0.4938 D $\lambda$ Now for interference Path difference at P.

{{dy} \over D}

= 4.8 $\lambda$ path difference at Q

{{dy} \over D}

= 9.6 $\lambda$ So orders of maxima in between P & Q is 5, 6, 7, 8, 9 So 5 bright fringes all present between P & Q.