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Wave Optics

JEE Physics · 98 questions · Page 1 of 10 · Click an option or "Show Solution" to reveal answer

Q1
When an unpolarized light of intensity I0{{I_0}} is incident on a polarizing sheet, the intensity of the light which does not get transmitted is
A 14 I0{1 \over 4}\,{I_0}
B 12 I0{1 \over 2}\,{I_0}
C I0{I_0}
D zero
Correct Answer
Option B
Solution
I=I0cos⁡2θI = {I_0}{\cos ^2}\theta

Intensity of polarized light

=I02= {{{I_0}} \over 2}

⇒\Rightarrow Intensity of untransmitted light

=I0−I02=I02= {I_0} - {{{I_0}} \over 2} = {{{I_0}} \over 2}
Q2
A beam of unpolarised light of intensity I0{{\rm I}_0} is passed through a polaroid AA and then through another polaroid BB which is oriented so that its principal plane makes an angle of 45∘{45^ \circ } relative to that of AA. The intensity of the emergent light is
A I0{{\rm I}_0}
B I02{{{I_0}} \over 2}
C I04{{{I_0}} \over 4}
D I08{{{I_0}} \over 8}
Correct Answer
Option C
Solution

Relation between intensities

Ir=(I02)cos⁡2(45∘)=I02×12=I04{{\rm I}_r} = \left( {{{{{\rm I}_0}} \over 2}} \right){\cos ^2}\left( {{{45}^ \circ }} \right) = {{{{\rm I}_0}} \over 2} \times {1 \over 2} = {{{{\rm I}_0}} \over 4}
Q3
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
Imax⁡=(I1+I2)2{I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}

= 4 I0

Q4
The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π\pi/2 at point P and π\pi at point Q. Then the difference between the resultant intensities at P and Q will be :
A 2 I
B 6 I
C 5 I
D 7 I
Correct Answer
Option B
Solution
IP=I+9I+2I×9Icos⁡π2=10I{I_P} = I + 9I + 2\sqrt {I \times 9I} \cos {\pi \over 2} = 10I
IQ=I+9I+2I×9Icos⁡π=4I{I_Q} = I + 9I + 2\sqrt {I \times 9I} \cos \pi = 4I

So,

IP−IQ=6I{I_P} - {I_Q} = 6I
Q5
When unpolarized light is incident at an angle of 60∘60^{\circ} on a transparent medium from air, the reflected ray is completely polarized. The angle of refraction in the medium is:
A 60∘60^{\circ}
B 90∘90^{\circ}
C 30∘30^{\circ}
D 45∘45^{\circ}
Correct Answer
Option C
Solution

By Brewster's law At complete reflection refracted ray and reflected ray are perpendicular.

Q6
The diffraction pattern of a light of wavelength 400 nm400 \mathrm{~nm} diffracting from a slit of width 0.2 mm0.2 \mathrm{~mm} is focused on the focal plane of a convex lens of focal length 100 cm100 \mathrm{~cm}. The width of the 1st 1^{\text{st }} secondary maxima will be :
A 2 mm
B 0.2 mm
C 0.02 mm
D 2 cm
Correct Answer
Option A
Solution

Width of

1st 1^{\text{st }}

secondary maxima

=λa⋅D=\frac{\lambda}{a} \cdot D

Here

a=0.2×10−3 mλ=400×10−9 mD=100×10−2\begin{aligned} & a=0.2 \times 10^{-3} \mathrm{~m} \\ & \lambda=400 \times 10^{-9} \mathrm{~m} \\ & D=100 \times 10^{-2} \end{aligned}

Width of

1st 1^{\text{st }}

secondary maxima

=400×10−90.2×10−3×100×10−2=2 mm\begin{aligned} & =\frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} \times 100 \times 10^{-2} \\ & =2 \mathrm{~mm} \end{aligned}
Q7
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 1:11: 1
B 4:14: 1
C 16:116: 1
D 9:19: 1
Correct Answer
Option D
Solution
Imax⁡=(4I0+I0)2=(3I0)2=9I0Imin⁡=(4I0−I0)2=I0Imax⁡Imin⁡=9:1\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}
Q8
The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index nn) is :
A tan⁡−1(1/n){\tan ^{ - 1}}\left( {1/n} \right)
B sin⁡−1(1/n){\sin ^{ - 1}}\left( {1/n} \right)
C sin⁡−1(n){\sin ^{ - 1}}\left( n \right)
D tan⁡−1(n){\tan ^{ - 1}}\left( n \right)
Correct Answer
Option D
Solution

The angle of incidence for total polarization is given by

tan⁡θ=n⇒θ=tan⁡−1n\tan \theta = n \Rightarrow \theta = {\tan ^{ - 1}}n

Where

nn

is the refractive index of the glass.

Q9
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is :
A three
B five
C infinite
D zero
Correct Answer
Option B
Solution

For constructive interference

d sin⁡θ=nλd\,\sin \theta = n\lambda

Given

d=2λ⇒sin⁡θ=n2d = 2\lambda \Rightarrow \sin \theta = {n \over 2}
n=0,1,−1,2,−2n = 0,1, - 1,2, - 2

hence five maxima are possible.

Q10
Two point white dots are 11 mmmm apart on a black paper. They are viewed by eye of pupil diameter 33 mm.mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light =500=500 nmnm ]
A 1m1m
B 5m5m
C 3m3m
D 6m6m
Correct Answer
Option B
Solution
yD≥1.22λd{y \over D} \ge 1.22{\lambda \over d}
⇒D≤yd(1.22)λ\Rightarrow D \le {{yd} \over {\left( {1.22} \right)\lambda }}
=10−3×3×10−3(1.22)×5×10−7= {{{{10}^{ - 3}} \times 3 \times {{10}^{ - 3}}} \over {\left( {1.22} \right) \times 5 \times {{10}^{ - 7}}}}
=3061≈5m= {{30} \over {61}} \approx 5m

∴\therefore

Dmax⁡=5m{D_{\max }} = 5m
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