Wave Optics — Page 3 | NEET Physics

Q21

A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aparture is illuminated normally by a parallel beam of wavelength 5

\times

10

-

5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

A 0.10 cm

B 0.25 cm

C 0.20 cm

D 0.15 cm

Correct Answer

Option D

Solution

Here, a = 0.02 cm = 2 × 10 –4 m $\lambda$ = 5 × 10 –5 cm = 5 × 10 –7 m D = 60 cm = 0.6 m Position of first minima on the diffraction pattern, y 1 =

{{D\lambda } \over a}

=

{{0.6 \times 5 \times {{10}^{ - 7}}} \over {2 \times {{10}^{ - 4}}}}

= 15 $\times$ 10 -4 m = 0.15 cm

Q22

The intensity at the maximum in a Young's double slit experiment is

I

0 . Distance between two slits is d = 5

\lambda

, where

\lambda

is the wavelength of light used in the expreriment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?

A

{3 \over 4}{I_0}

B

{{{I_0}} \over 2}

C I 0

D

{{{I_0}} \over 4}

Correct Answer

Option B

Solution

Path difference S 2 P – S 1 P =

\sqrt {{{\left( {50\lambda } \right)}^2} + {{\left( {5\lambda } \right)}^2}}

- 50 $\lambda$ = 0.25 $\lambda$ S 2 P – S 1 P =

{\lambda \over 4}

Phase difference,

\Delta

$\phi$ =

{{2\pi } \over \lambda } \times {\lambda \over 4}

=

{\pi \over 2}

So, resultant intensity at the desired point 'P' is I = I 0 cos 2

{\phi \over 2}

= I 0 cos 2

{\pi \over 4}

=

{{{I_0}} \over 2}

Q23

In a diffraction pattern due to a single slit of width

a

, the first minimum is observed at an angle 30 o when light of wavelength 5000

\mathop A\limits^ \circ

is incident on the slit. The first secondary maximum is observed at an angle of

A sin

-

1

\left( {{1 \over 2}} \right)

B

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

C

{\sin ^{ - 1}}\left( {{1 \over 4}} \right)

D

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

Correct Answer

Option B

Solution

For the first minima, the path difference between extreme waves sin $\theta$ =

{\lambda \over a}

$\Rightarrow$

{1 \over 2}

=

{\lambda \over a}

$\Rightarrow$

a

= 2 $\lambda$ For first secondary maximum, the path difference between extreme waves

\sin \theta ' = {{3\lambda } \over {2a}}

=

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

$\Rightarrow$

\theta '

=

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

Q24

At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

A

\pi

radian

B

{\pi \over 8}

radian

C

{\pi \over 4}

radian

D

{\pi \over 2}

radian

Correct Answer

Option A

Solution

For first minima at P AP – BP = $\lambda$ AP – MP =

{\lambda \over 2}

So phase difference, $\phi$ =

{{2\pi } \over \lambda } \times {\lambda \over 2}

= $\pi$ radian

Q25

Two slits in Young's experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern,

{{{I_{max}}} \over {{I_{min}}}}

is

A

{{49} \over {121}}

B

{4 \over 9}

C

{9 \over 4}

D

{{121} \over {49}}

Correct Answer

Option C

Solution

Given, The ratio of slits width =

{1 \over {25}}

$\therefore$

{{{I_1}} \over {{I_2}}} = {{25} \over 1}

I $\propto$ A 2 $\Rightarrow$

{{{I_1}} \over {{I_2}}} = {{A_1^2} \over {A_2^2}}

=

{{25} \over 1}

$\Rightarrow$

{{{A_1}} \over {{A_2}}} = {5 \over 1}

{{{A_{\max }}} \over {{A_{\min }}}} = {{{A_1} + {A_2}} \over {{A_1} - {A_2}}}

=

{{5 + 1} \over {5 - 1}}

=

{6 \over 4}

=

{3 \over 2}

$\therefore$

{{{I_{\max }}} \over {{I_{\min }}}} = {{A_{\max }^2} \over {A_{\min }^2}}

=

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

=

{9 \over 4}

Q26

For a parallel beam of monochromatic light of wavelength '

\lambda

' , diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the wifth of the central maxima will be

A

{{Da} \over \lambda }

B

{{2Da} \over \lambda }

C

{{2D\lambda } \over a}

D

{{D\lambda } \over a}

Correct Answer

Option C

Solution

For central maxima, sin $\theta$ =

{\lambda \over a}

Also, $\theta$ is very-very small so sin $\theta$ $\approx$ tan $\theta$ =

{y \over D}

$\therefore$

{y \over D}

=

{\lambda \over a}

$\Rightarrow$ y =

{{\lambda D} \over a}

Width of central maxima = 2y =

{{2D\lambda } \over a}

Q27

In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern ?

A 0.5 mm

B 0.02 mm

C 0.2 mm

D 0.1 mm

Correct Answer

Option C

Solution

Fringe width, $\beta$ =

{{D\lambda } \over d}

As per question, width of central maxima of single slit pattern = width of 10 maxima of double slit pattern

{{D2\lambda } \over a} = 10\left( {{{\lambda D} \over d}} \right)

$\Rightarrow$

a

=

{{2d} \over {10}}

=

{{2 \times {{10}^{ - 3}}} \over {10}}

= 0.2 $\times$ 10 -3 m = 0.2 mm

Q28

A beam of light of

\lambda = 600

nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is

A 1.2 cm

B 1.2 mm

C 2.4 cm

D 2.4 mm

Correct Answer

Option D

Solution

Distance between the first dark fringes on either side of the central bright fringe is also the width of central maximum.

Width of central maximum =

{{{2\lambda D} \over a}}

=

{{2 \times 600 \times {{10}^{ - 9}} \times 2} \over {{{10}^{ - 3}}}}

= 24 × 10 –4 m = 2.4 × 10 –3 m = 2.4 mm

Q29

In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference

\lambda

is K, (

\lambda

being the wavelength of light used). The intensity at a point where the path difference is

\lambda

/4 will be

A K

B K/4

C K/2

D zero

Correct Answer

Option C

Solution

Phase difference, $\phi$ =

{{2\pi } \over \lambda }

$\times$ Path difference When path difference is $\lambda$ , then $\phi$ =

{{2\pi } \over \lambda } \times \lambda

= 2 $\pi$ $\therefore$ I = 4I 0

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

= 4I 0 cos 2 ( $\pi$ ) = 4I 0 = K ....(1) When path difference is ,

{\lambda \over 4}

then $\phi$ =

{{2\pi } \over \lambda } \times {\lambda \over 4}

=

{{\pi \over 2}}

$\therefore$ I = 4I 0

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

= 2I 0 =

{K \over 2}

Q30

The reddish appearance of the sun at sunrise and sunset is due to

A the scattering of light

B the polarisation of light

C the colour of the sun

D the colour of the sky

Correct Answer

Option A

Solution

The reddish appearance of the sun at sunrise and sunset is due to the scattering of light.