Wave Optics — Page 10 | JEE Physics

Q91

Visible light of wavelength 6000

\times

10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at

\theta

1, then

\theta

1, is close to :

A 45o

B 30o

C 25o

D 20o

Correct Answer

Option C

Solution

For 2nd minima

\sin \theta = {{2\lambda } \over d }

$\Rightarrow$

\sin 60^\circ = {{2\lambda } \over d}

$\Rightarrow$

{\lambda \over d} = {{\sqrt 3 } \over 4}

For 1st minima

\sin {\theta _1} = {\lambda \over d}

=

{{\sqrt 3 } \over 4}

$\Rightarrow$

{\theta _1} =

25o

Q92

In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle

{1 \over {40}}

by using light of wavelength

\lambda

1. When the light of wavelength

\lambda

2 is used a bright fringe is seen at the same angle in the same set up. Given that

\lambda

1 and

\lambda

2 are in visible range (380 nm to 740 nm), their values are -

A 400 nm, 500 nm

B 625 nm, 500 nm

C 380 nm, 500 nm

D 380 nm, 525 nm

Correct Answer

Option B

Solution

Path difference = d sin $\theta$ $\approx$ d $\theta$ = 0.1 $\times$

{1 \over {40}}

mm = 2500nm or bright fringe, path difference must be integral multiple of $\lambda$ . $\therefore$ 2500 = n $\lambda$ 1 = m $\lambda$ 2 $\therefore$ $\lambda$ 1 = 625 (from n = 4), $\lambda$ 2 = 500 (from m = 5)

Q93

Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600nm. coming from a distant object, the limit of resolution of the telescope is close to :-

A 3.0 × 10–7 rad

B 4.5 × 10–7 rad

C 1.5 × 10–7 rad

D 2.0 × 10–7 rad

Correct Answer

Option A

Solution

Limit of resolution =

{{1.22\lambda } \over d}

=

{{1.22 \times 600 \times {{10}^{ - 9}}} \over {250 \times {{10}^{ - 2}}}}

= 2.9 × 10–7 rad.

Q94

'

n

' polarizing sheets are arranged such that each makes an angle

45^{\circ}

with the preceeding sheet. An unpolarized light of intensity I is incident into this arrangement. The output intensity is found to be

I / 64

. The value of

n

will be:

A 4

B 5

C 3

D 6

Correct Answer

Option D

Solution

After passing through first sheet

I_1=\frac{I}{2}

After passing through second sheet

I_2=I_1 \cos ^2\left(45^{\circ}\right)=\frac{I}{4}

After passing through $n^{\text{th }}$ sheet

\begin{aligned} & I_{\mathrm{n}}=\frac{I}{2^{\mathrm{n}}}=\frac{I}{64} \\\\ & n=6 \end{aligned}

Q95

An initially parallel cylindrical beam travels in a medium of refractive index

\mu \left( I \right) = {\mu _0} + {\mu _2}\,I,

where

{\mu _0}

and

{\mu _2}

are positive constants and

I

is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. The initial shape of the wavefront of the beam is

A convex

B concave

C convex near the axis and concave near the periphery

D planar

Correct Answer

Option D

Solution

Initially the parallel beam is cylindrical. Therefore, the wave-front will be planar.

Q96

In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is

A 15.6 mm

B 1.56 mm

C 7.8 mm

D 9.75 mm

Correct Answer

Option C

Solution

Let n1th fringe formed due to first wavelength and n2th fringe formed due to second wavelength coincide.

So their distance from common central maxima will be same. yn1 = yn2

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

$\Rightarrow$

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

=

{{520 \times {{10}^{ - 9}}} \over {650 \times {{10}^{ - 9}}}} = {4 \over 5}

Hence, distance of the point of coincidence from the central maxima is y =

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

=

{{4 \times 450 \times {{10}^{ - 9}} \times 15} \over {0.5 \times {{10}^{ - 3}}}}

= 7.8 mm

Q97

For a specific wavelength 670 nm of light coming from a galaxy moving with velocity v, the observed wavelength is 670.7 nm. The value of v is :

A 3

\times

108 ms

-

1

B 3

\times

1010 ms

-

1

C 3.13

\times

105 ms

-

1

D 4.48

\times

105 ms

-

1

Correct Answer

Option C

Solution

{\lambda _{obs}} = {\lambda _{source}}\sqrt {{{1 + {v \over C}} \over {1 - {v \over C}}}}

For

v

{{670.7} \over {670}} = 1 + {v \over C}

\Rightarrow v = {{0.7} \over {670}} \times 3 \times {10^8}

m/s

\Rightarrow v \simeq 3.13 \times {10^5}$$ m/s

Q98

Two monochromatic light beams have intensities in the ratio 1:9. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is

A

8: 1

B

4: 1

C

3: 1

D

9: 1

Correct Answer

Option B

Solution

To find the ratio of the intensities of the maximum to the minimum in an interference pattern formed by two monochromatic light beams with intensity ratios of 1:9, you use the formula: $\dfrac{I_{\max }}{I_{\min }} = \dfrac{\left(\sqrt{I_1} + \sqrt{I_2}\right)^2}{\left(\sqrt{I_1} - \sqrt{I_2}\right)^2}$ Given that the intensity ratio is 1:9, we have $I_1 = 1$ and $I_2 = 9$ .

Calculate $\sqrt{I_1}$ and $\sqrt{I_2}$ : $\sqrt{I_1} = \sqrt{1} = 1$ $\sqrt{I_2} = \sqrt{9} = 3$ Substitute these values into the formula: $\dfrac{I_{\max }}{I_{\min }} = \dfrac{(1 + 3)^2}{(1 - 3)^2} = \dfrac{4^2}{2^2} = \dfrac{16}{4} = 4$ Thus, the ratio of the intensities of the maximum to the minimum is 4:1.