Wave Optics — Page 3 | JEE Physics

Q21

In an interference experiment the ratio of amplitudes of coherent waves is

{{{a_1}} \over {{a_2}}} = {1 \over 3}

. The ratio of maximum and minimum intensities of fringes will be :

A 2

B 4

C 18

D 9

Correct Answer

Option B

Solution

{{{I_{\max }}} \over {{I_{\min }}}} = {{{{\left( {{a_1} + {a_2}} \right)}^2}} \over {{{\left( {{a_1} - {a_2}} \right)}^2}}} = {{{{\left( {1 + 3} \right)}^2}} \over {{{\left( {1 - 3} \right)}^2}}} = {{16} \over 4}

= 4

Q22

If the source of light used in a Young's double slit experiment is changed from red to violet :

A the fringes will become brighter.

B the intensity of minima will increase.

C consecutive fringe lines will come closer.

D the central bright fringe will become a dark fringe.

Correct Answer

Option C

Solution

According to Young's double slit experiment, the distance of nth bright fringe from the centre,

{y_n} = {{n\lambda D} \over d}

Since,

{\lambda _{violet}} < {\lambda _{red}}

$\therefore$

{y_{violet}} < {y_{red}}

$\therefore$ Consecutive fringe lines will come closer.

Q23

In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of a wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be

A 28

B 24

C 30

D 18

Correct Answer

Option A

Solution

Let the length of segment is "

l

" Let N is the no. of fringes in "

l

" and w is fringe width. $\therefore$ Nw =

l

$\Rightarrow$ N

\left( {{{\lambda D} \over d}} \right)

=

l

As in both cases segment length is same. $\therefore$

{{{{N_1}{\lambda _1}D} \over d} = l}

and

{{{{N_2}{\lambda _2}D} \over d} = l}

$\therefore$

{{{{N_1}{\lambda _1}D} \over d}}

=

{{{{N_2}{\lambda _2}D} \over d}}

$\Rightarrow$

{{N_1}{\lambda _1}}

=

{{N_2}{\lambda _2}}

$\Rightarrow$ 16 × 700 = N2 × 400 $\Rightarrow$ N2 = 28

Q24

In a double slit experiment, at a certain point on the screen the path difference between the two interfering waves is

{1 \over 8}

th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is :

A 0.853

B 0.568

C 0.672

D 0.760

Correct Answer

Option A

Solution

\Delta

X =

{\lambda \over 8}

\Delta

$\phi$ =

{{2\pi } \over \lambda }

\Delta

X =

{{2\pi } \over \lambda }{\lambda \over 8}

=

{\pi \over 4}

I = I0

{\cos ^2}\left( {{{\Delta \phi } \over 2}} \right)

$\Rightarrow$

{I \over {{I_0}}}

=

{\cos ^2}\left( {{{{\pi \over 4}} \over 2}} \right)

$\Rightarrow$

{I \over {{I_0}}}

=

{\cos ^2}\left( {{\pi \over 8}} \right)

= 0.853

Q25

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. 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.

A 4 : 1

B 2 : 1

C 1 : 4

D 3 : 1

Correct Answer

Option A

Solution

Given, amplitude $\propto$ width of slit $\Rightarrow$ A2 = 3A1 We know that,

{{{I_{\max }}} \over {{I_{\min }}}} = {{{{(\sqrt {{I_1}} + \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {{I_1}} - \sqrt {{I_2}} )}^2}}}

$\because$ Intensity, I $\propto$ A2 $\therefore$

{{{I_{\max }}} \over {{I_{\min }}}} = {{{{({A_1} + {A_2})}^2}} \over {{{(|{A_1} - {A_2}|)}^2}}}

= {\left( {{{{A_1} + 3{A_1}} \over {|{A_1} - 3{A_1}|)}}} \right)^2} = {\left( {{{4{A_1}} \over {2{A_1}}}} \right)^2} = {4 \over 1}

$\therefore$

{I_{\max }}:{I_{\min }} = 4:1

Q26

Two coherent light sources having intensity in the ratio 2x produce an interference pattern. The ratio

{{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}}

will be :

A

{{2\sqrt {2x} } \over {2x + 1}}

B

{{2\sqrt {2x} } \over {x + 1}}

C

{{\sqrt {2x} } \over {x + 1}}

D

{{\sqrt {2x} } \over {2x + 1}}

Correct Answer

Option A

Solution

Given,

{{{I_1}} \over {{I_2}}} = 2x

We know,

{{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}

= {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}

= {\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)^2}

Now,

{{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}}

= {{{{{I_{\max }}} \over {{I_{\min }}}} - 1} \over {{{{I_{\max }}} \over {{I_{\min }}}} + 1}}

= {{{{\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)}^2} - 1} \over {{{\left( {{{\sqrt {2x} + 1} \over {\sqrt {2x} - 1}}} \right)}^2} + 1}}

= {{{{\left( {\sqrt {2x} + 1} \right)}^2} - {{\left( {\sqrt {2x} - 1} \right)}^2}} \over {{{\left( {\sqrt {2x} + 1} \right)}^2} + {{\left( {\sqrt {2x} - 1} \right)}^2}}}

= {{4\sqrt {2x} } \over {2 + 4x}} = {{2\sqrt {2x} } \over {1 + 2x}}

Q27

Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter 0.1

\mu

m. If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that:

A its size increases, but intensity decreases

B its size increases, and intensity increases

C its size decreases, but intensity increases

D its sizes decreases, and intensity decreases

Correct Answer

Option C

Solution

Yes, you're absolutely correct.

This equation you provided is for the angular size of the central maximum (or central diffraction disk) in a circular aperture diffraction pattern (like a pinhole) :

\sin \theta = \frac{1.22\lambda}{D}

where : θ is the angle subtended by the radius of the central maximum at the pinhole, λ is the wavelength of the light, and D is the diameter of the aperture or pinhole.

If D (the diameter of the pinhole) is increased, then sinθ (and hence θ itself, for small θ) will decrease, implying that the size of the central maximum or diffraction disk will decrease.

This is because less diffraction (bending of light) occurs when the pinhole is larger.

At the same time, increasing the size of the pinhole allows more light to pass through, which increases the intensity (brightness) of the light in the diffraction pattern.

So, increasing the diameter of the pinhole decreases the size of the diffraction pattern but increases its intensity, which confirms Option C.

Q28

In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be :

A 0.50 mm

B 0.25 mm

C 1 mm

D 0.75 mm

Correct Answer

Option B

Solution

Fringe width ( $\beta$ ) =

{{\lambda D} \over d}

d = 2 $\times$ 10 $-$ 3 m $\lambda$ = 500 $\times$ 10 $-$ 9 m D = 1 m Now $\beta$ =

{{500 \times {{10}^{ - 9}} \times 1} \over {2 \times {{10}^{ - 3}}}}

$\beta$ =

{5 \over 2} \times {10^{ - 4}}

$\beta$ = 2.5 $\times$ 10 $-$ 4 $\beta$ = 0.25 mm

Q29

In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890

\mathop A\limits^o

is :-

A 1178

\times

10

-

9 m

B 1178

\times

10

-

6 m

C 1178

\times

10

-

12 m

D 5890

\times

10

-

7 m

Correct Answer

Option B

Solution

In the double-slit experiment, the position of a bright fringe is given by the formula : $y = \dfrac{{m\lambda L}}{{d}}$ , where : m is the order of the fringe (1 for the first bright fringe, 2 for the second, etc.) $\lambda$ is the wavelength of the light (in meters) L is the distance from the slits to the screen (in meters) d is the distance between the slits (in meters) We need to find the difference in position between the first and third bright fringes.

So, we find the position of both and subtract the position of the first from the position of the third : $y_3 = \dfrac{{3\lambda L}}{{d}}$ $y_1 = \dfrac{{\lambda L}}{{d}}$ Then, the difference between the third and the first bright fringes is : $y_3 - y_1 = 2\dfrac{{\lambda L}}{{d}}$ Now, let's plug the given values: $\lambda = 5890 \,Å = 5890 \times 10^{-10} \, m$ (since 1 Å = $10^{-10}$ meters), L = 0.5 m, and d = 0.5 mm = 0.5 $\times 10^{-3}$ m :

y_3 - y_1 = 2 \times \frac{5890 \times 10^{-10} \, \text{m} \times 0.5 \, \text{m}}{0.5 \times 10^{-3} \, \text{m}} = 1178 \times 10^{-6} \, \text{m}

So, the correct answer is 1178 $\times 10^{-6}$ m, which corresponds to Option B.

Q30

In the Young's double slit experiment, the distance between the slits varies in time as d(t) = d0 + a0 sin

\omega

t; where d0,

\omega

and a0 are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

A

{{2\lambda D({d_0})} \over {(d_0^2 - a_0^2)}}

B

{{2\lambda D{a_0}} \over {(d_0^2 - a_0^2)}}

C

{{\lambda D} \over {d_0^2}}{a_0}

D

{{\lambda D} \over {{d_0} + {a_0}}}

Correct Answer

Option B

Solution

Fringe Width,

\beta = {{\lambda D} \over d}

{\beta _{\max }} \Rightarrow {d_{\min }}

and

{\beta _{\min }} \Rightarrow {d_{\max }}

d = {d_0} + {a_0}\sin \omega t

{d_{\max }} = {d_0} + {a_0}

and

{d_{\min }} = {d_0} - {a_0}

$\therefore$

{\beta _{\min }} = {{\lambda D} \over {{d_0} + {a_0}}}

and $\therefore$

{\beta _{\max }} = {{\lambda D} \over {{d_0} - {a_0}}}

{\beta _{\max }} - {\beta _{\min }} = {{\lambda D} \over {{d_0} - {a_0}}} - {{\lambda D} \over {{d_0} + {a_0}}} = {{2\lambda D{a_0}} \over {d_0^2 - a_0^2}}