Wave Optics — Page 4 | NEET Physics

Q31

A parallel beam of light of wavelength

\lambda

is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpenficular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is

A 2

\pi

B 3

\pi

C 4

\pi

D

\pi

\lambda

Correct Answer

Option C

Solution

\Delta

$\phi$ =

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

Path Difference =

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

= 4 $\pi$

Q32

In Young's double slit experiment the distance between the slits and the screen is doubled. The separation between the slits is reduced to half. As a result the fringe width

A is halved

B becomes four times

C remains unchanged

D is doubled

Correct Answer

Option B

Solution

Fringe width, $\beta$ =

{{\lambda D} \over d}

From question D' = 2D and d' =

{d \over 2}

$\therefore$ $\beta$ ' =

{{\lambda D'} \over {d'}}

= 4 $\beta$

Q33

In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths

{\lambda _1}

= 12000

\mathop A\limits^ \circ

and

{\lambda _2}

= 10000

\mathop A\limits^ \circ

. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other ?

A 4 mm

B 3 m

C 8 mm

D 6 mm

Correct Answer

Option D

Solution

From the question, n 1 $\lambda$ 1 = n 2 $\lambda$ 2 So,

{{{n_1}} \over {{n_2}}} = {{{\lambda _2}} \over {{\lambda _1}}}

=

{{10000} \over {12000}} = {5 \over 6}

Hence, minima n 1 and n 2 are 5 and 6. X min =

{{{n_1}{\lambda _1}D} \over d} = {{5\left( {12000 \times {{10}^{ - 10}}} \right)2} \over {2 \times {{10}^{ - 3}}}}

= 6 × 10 –3 m = 6 mm

Q34

The frequency of a light wave in a material is 2

\times

10 14 Hz and wavelength is 5000

\mathop A\limits^ \circ

. The refractive index of material will be

A 1.50

B 3.00

C 1.33

D 1.40

Correct Answer

Option B

Solution

By using v = n $\lambda$ $\therefore$ v = 2 $\times$ 10 14 $\times$ 5000 $\times$ 10 -10 = 10 8 m/s Refractive index of the material, $\mu$ =

{c \over v}

=

{{3 \times {{10}^8}} \over {{{10}^8}}}

= 3

Q35

The angular resolution of a 10 cm diameter telescope at a wavelength of 5000

\mathop A\limits^ \circ

is of the order of

A 10 6 rad

B 10

-

2 rad

C 10

-

4 rad

D 10

-

6 rad.

Correct Answer

Option C

Solution

The angular resolution,

\Delta

$\theta$ =

{{1.22\lambda } \over D}

=

{{1.22 \times 5000 \times {{10}^{ - 8}}} \over {0.1}}

= 6.1 $\times$ 10 -4 $\therefore$ Order = 10 -4

Q36

A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometer from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000

\mathop A\limits^ \circ

, is of the order of

A 0.5 m

B 5 m

C 5 mm

D 5 cm

Correct Answer

Option C

Solution

Here,

{x \over {1000}} = {{1.22\lambda } \over D}

$\Rightarrow$ x =

{{1.22 \times 5 \times {{10}^3} \times {{10}^{ - 10}} \times {{10}^3}} \over {10 \times {{10}^{ - 2}}}}

$\Rightarrow$ x = 1.22 $\times$ 5 $\times$ 10 -3 m = 6.1 mm $\therefore$ x is of the order of 5 mm.

Q37

A ray of light travelling in air have wavelength

\lambda

, frequency n, velocity v and intensity

I

. If this ray enters into water then these parameters are

\lambda

', n', v' and

I

' respectively. Which relation is correct from following ?

A

\lambda

=

\lambda

'

B n = n'

C v = v'

D

I

=

I

'.

Correct Answer

Option B

Solution

Frequency remains same.

Q38

When an unpolarized light of intensity

{{I_0}}

is incident on a polarizing sheet, the intensity of the light which does not get transmitted is

A

{1 \over 4}\,{I_0}

B

{1 \over 2}\,{I_0}

C

{I_0}

D zero

Correct Answer

Option B

Solution

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

Intensity of polarized light

= {{{I_0}} \over 2}

$\Rightarrow$ Intensity of untransmitted light

= {I_0} - {{{I_0}} \over 2} = {{{I_0}} \over 2}

Q39

The light waves from two coherent sources have same intensity I1 = I2 = I0. In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima?

A I0

B 2 I0

C 5 I0

D 4 I0

Correct Answer

Option D

Solution

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

= 4 I0

Q40

The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is:

A

1: 1

B

4: 1

C

16: 1

D

9: 1

Correct Answer

Option D

Solution

\begin{aligned} & I_{\max }=\left(\sqrt{4 I_0}+\sqrt{I_0}\right)^2=\left(3 \sqrt{I_0}\right)^2=9 I_0 \\ & I_{\min }=\left(\sqrt{4 I_0}-\sqrt{I_0}\right)^2=I_0 \\ & \frac{I_{\max }}{I_{\min }}=9: 1 \end{aligned}