Wave Optics — Page 6 | JEE Physics

Q51

Two plane polarized light waves combine at a certain point whose electric field components are

\begin{aligned} & E_1=E_0 \operatorname{Sin} \omega t \\ & E_2=E_0 \operatorname{Sin}\left(\omega t+\frac{\pi}{3}\right) \end{aligned}

Find the amplitude of the resultant wave.

A

1.7 \mathrm{~E}_0

B

\mathrm{E}_0

C

0.9 \mathrm{~E}_0

D

3.4 \mathrm{~E}_0

Correct Answer

Option A

Solution

To find the amplitude of the resultant wave for the given electric field components, we use the formula for the resultant amplitude of two combining waves.

These components are: $E_1 = E_0 \sin \omega t$ $E_2 = E_0 \sin \left(\omega t + \dfrac{\pi}{3}\right)$ The general formula for the amplitude $E$ of two superimposed waves $E_1$ and $E_2$ with a phase difference is: $E = \sqrt{E_1^2 + E_2^2 + 2E_1E_2 \cos\phi}$ where $\phi$ is the phase difference between the two waves.

In this scenario, $\phi = \dfrac{\pi}{3}$ .

Substituting the given expressions and phase difference into the formula, we get: $E = \sqrt{(E_0)^2 + (E_0)^2 + 2(E_0)(E_0)\cos\left(\dfrac{\pi}{3}\right)}$ Given that $\cos \left( \dfrac{\pi}{3} \right) = \dfrac{1}{2}$ , this simplifies to: $E = \sqrt{2E_0^2 + E_0^2} = \sqrt{3}E_0$ Thus, the amplitude of the resultant wave is $\sqrt{3}E_0 \approx 1.73E_0$ .

Q52

Two polarisers

P_1

and

P_2

are placed in such a way that the intensity of the transmitted light will be zero. A third polariser

P_3

is inserted in between

P_1

and

P_2

, at particular angle between

P_2

and

P_3

. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers

P_2

and

P_3

is :

A

\pi / 6

B

\dfrac{\pi}{3}

C

\dfrac{\pi}{4}

D

\pi / 8

Correct Answer

Option C

Solution

Through $P_2 I_1=I_0 \sin ^2\left(\dfrac{\pi}{2}-\theta\right)$

I_1=I_0 \cos ^2 \theta

Through $P_3 I_{\text{net }}=\left(I_0 \cos ^2 \theta\right) \sin ^2 \theta$

I_{\text{net }}=\frac{I_0}{4}[\sin (2 \theta)]^2 \text{ for max } I_{\text{net }} \theta=45^{\circ}

So angle between $\mathrm{P}_2$ and $\mathrm{P}_3=\dfrac{\pi}{4}$ Correct Ans. (1)

Q53

In a Young's double slit experiment, the slits are separated by 0.2 mm . If the slits separation is increased to 0.4 mm , the percentage change of the fringe width is :

A

25 \%

B

50 \%

C

0 \%

D

100 \%

Correct Answer

Option B

Solution

In Young's double slit experiment, the fringe width $\beta$ is given by:

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

where: $\lambda$ is the wavelength of light,

D

is the distance from the slits to the screen,

d

is the separation between the slits. Here's how the change affects the fringe width: Initial Situation: When

d = 0.2 \text{ mm}

, the fringe width is:

\beta_{\text{initial}} = \frac{\lambda D}{0.2 \text{ mm}}

After Increasing Slit Separation: When

d

is increased to

0.4 \text{ mm}

:

\beta_{\text{new}} = \frac{\lambda D}{0.4 \text{ mm}}

Comparing the Two Fringe Widths: Notice that:

\beta_{\text{new}} = \frac{1}{2} \times \frac{\lambda D}{0.2 \text{ mm}} = \frac{1}{2} \beta_{\text{initial}}

This means the fringe width is halved, which is a reduction of

50\%

. Thus, the fringe width decreases by

50\%

when the slit separation is increased from

0.2 \text{ mm}

to

0.4 \text{ mm}

. The correct answer is Option B:

50\%

.

Q54

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

A

3: 1

B

(2 \sqrt{2}+1):(2 \sqrt{2}-1)

C

9: 1

D

(3+2 \sqrt{2}):(3-2 \sqrt{2})

Correct Answer

Option D

Solution

In Young's double-slit interference experiment, the width of one slit is half that of the other.

The intensity, $\mathrm{I}$ , is proportional to the slit width.

This gives us: $\begin{align*} \mathrm{I}_1 &= \mathrm{I}_0, \\ \mathrm{I}_2 &= 2 \mathrm{I}_0. \end{align*}$ The maximum intensity, $\mathrm{I}_{\max}$ , occurs when the amplitudes add constructively: $\mathrm{I}_{\max} = \left(\sqrt{\mathrm{I}_1} + \sqrt{\mathrm{I}_2}\right)^2.$ The minimum intensity, $\mathrm{I}_{\min}$ , occurs when the amplitudes interfere destructively: $\mathrm{I}_{\min} = \left(\sqrt{\mathrm{I}_1} - \sqrt{\mathrm{I}_2}\right)^2.$ Now, using $\mathrm{I}_1 = \mathrm{I}_0$ and $\mathrm{I}_2 = 2 \mathrm{I}_0$ , we find: $\dfrac{\mathrm{I}_{\max}}{\mathrm{I}_{\min}} = \dfrac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} = \dfrac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}.$

Q55

A monochromatic light of frequency

5 \times 10^{14} \mathrm{~Hz}

travelling through air, is incident on a medium of refractive index ' 2 '. Wavelength of the refracted light will be :

A 400 nm

B 300 nm

C 600 nm

D 500 nm

Correct Answer

Option B

Solution

To find the wavelength of the refracted light in a medium with a refractive index of 2, we can use the relationship between frequency, wavelength, and the speed of light.

The frequency ( $f$ ) of the light remains the same when it enters the medium.

The formula relating frequency, wavelength, and velocity is: $v = f \lambda$ Where: $v$ is the speed of light in the medium, $f$ is the frequency of the light, $\lambda$ is the wavelength of the light.

The wavelength of light in the medium ( $\lambda_{\text{medium}}$ ) can be found using the refractive index ( $\mu$ ) and the wavelength in a vacuum ( $\lambda_{\text{vacuum}}$ ) as follows: $\lambda_{\text{medium}} = \dfrac{\lambda_{\text{vacuum}}}{\mu}$ Given: Frequency of the light, $f = 5 \times 10^{14} \, \text{Hz}$ Speed of light in a vacuum, $v_{\text{vacuum}} = 3 \times 10^{8} \, \text{m/s}$ Refractive index of the medium, $\mu = 2$ First, calculate $\lambda_{\text{vacuum}}$ : $\lambda_{\text{vacuum}} = \dfrac{v_{\text{vacuum}}}{f} = \dfrac{3 \times 10^{8}}{5 \times 10^{14}}$ Now calculate $\lambda_{\text{medium}}$ : $\lambda_{\text{medium}} = \dfrac{\lambda_{\text{vacuum}}}{2} = \dfrac{3 \times 10^{8}}{2 \times 5 \times 10^{14}}$ $\lambda_{\text{medium}} = 0.3 \times 10^{-6} \, \text{m}$ Converting to nanometers: $0.3 \times 10^{-6} \, \text{m} = 300 \, \text{nm}$ Therefore, the wavelength of the refracted light in the medium is 300 nm.

Q56

In Young’s double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance d0 between the slits. If the angular resolution of the eye is

({{{1}} \over {60}})^o

, the value of d0 is close to :

A 1 mm

B 2 mm

C 4 mm

D 3 mm

Correct Answer

Option B

Solution

We know, Fringe width, $\beta$ =

{{\lambda D} \over {{d_0}}}

From image, $\theta$ =

{\beta \over D}

$\Rightarrow$ $\theta$ =

{\lambda \over {{d_0}}}

$\Rightarrow$ d0 =

{\lambda \over \theta }

=

{{600 \times {{10}^{ - 9}}} \over {{1 \over {60}} \times {\pi \over {180}}}}

= 2.06 $\times$ 10 $-$ 3

\simeq

2 mm

Q57

Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600 nm. To see the stars just resolved by the telescope, the minimum distance between them should be (1 light year = 9.46

\times

1015 m) of the order of :

A 106 km

B 108 km

C 1011 km

D 1010 km

Correct Answer

Option B

Solution

The limit of resolution of a telescope,

\Delta

$\theta$ =

{{1.22\,\,\lambda } \over D}

=

{l \over R}

$\therefore$

l

=

{{1.22\,\,\lambda R} \over D}

=

{{1.22 \times 6 \times {{10}^{ - 7}} \times 10 \times 9.46 \times 10{}^{15}} \over {30 \times {{10}^{ - 2}}}}

= 2.31 $\times$ 108 km

Q58

In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength

\lambda

= 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range

-

30o

\le

\theta

\le

30o is :

A 640

B 320

C 321

D 641

Correct Answer

Option D

Solution

We know, path difference, d sin $\theta$ = n $\lambda$ here n = no of bright fringer in the angle here given d = 0.32 $\times$ 10-3 m $\lambda$ = 500 $\times$ 10 $-$ 9 m $\therefore$ 0.32 $\times$ 10 $-$ 3 sin30o = n $\times$ 500 $\times$ 10 $-$ 9 $\Rightarrow$ n = 320 Total number of maxima in the range $-$ 30o $\theta$ $\le$ 30o is = 320 $\times$ 2 + 1 = 641

Q59

In a double slit experiment, when a thin film of thickness t having refractive index

\mu

. is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is (

\lambda

is the wavelength of the light used) :

A

{\lambda \over {2\left( {\mu - 1} \right)}}

B

{\lambda \over {\left( {2\mu - 1} \right)}}

C

{{2\lambda } \over {\left( {\mu - 1} \right)}}

D

{\lambda \over {\left( {\mu - 1} \right)}}

Correct Answer

Option D

Solution

As we know, Path difference introduced by thin film,

\Delta=(\mu-1) t

.......(i) and if fringe pattern shifts by one frings width, then path difference,

\Delta=1 \times \lambda=\lambda

.......(ii) So, from Eqs. (i) and (ii), we get

(\mu-1) t=\lambda

$\Rightarrow$

t=\frac{\lambda}{\mu-1}

Q60

With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890

\mathop A\limits^o

is observed at 5896

\mathop A\limits^o

?

A 306 km/sec

B 322 km/sec

C 296 km/sec

D 336 km/sec

Correct Answer

Option A

Solution

f = {f_0}\sqrt {{{1 + \beta } \over {1 - \beta }}}

\beta = {v \over c}

{f \over {{f_0}}} = \sqrt {{{1 + \beta } \over {1 - \beta }}}

{\left( {1 + {{\Delta f} \over {{f_0}}}} \right)^2} = (1 + \beta ){(1 - \beta )^{ - 1}}

$\beta$ is small compared to 1

\left( {1 + {{2\Delta f} \over {{f_0}}}} \right) = (1 + 2\beta )

\beta = {{\Delta f} \over {{f_0}}} = {v \over c}

v = 6 \times {c \over {5890}} = 305.6

km/s