Wave Optics Test

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Wave Optics Test - 39

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Question 1
1 / -0

The solar glare of sunlight bouncing off water or snow can be a real problem for drivers. The reflecting sunlight is horizontally polarized, meaning that the light waves oscillate at an angle of $$90^o$$ to a normal line drawn perpendicular to the Earth. At what angle relative to this normal line should sunglasses be polarized if they are to be effective against solar glare?

A
$$0^o$$

B
$$30^o$$

C
$$45^o$$

D
$$60^o$$

E
$$90^o$$

Solution

The idea behind polarized sunglasses is to eliminate the glare. If the solar glare is all at a $$90^o$$ angle to the normal line, sunglasses polarized at a $$ 0^o$$ angle to this normal will not allow any of the glare to pass. Most other light is not polarized, so it will still be possible to see the road and other cars, but the distracting glare will cease to be a problem.
Question 2
1 / -0

Find out correct option which describe the Red shift of distant galaxies.

A
Expansion of the universe

B
The Uncertainty Principle

C
Black holes

D
Dark matter

E
Superconductivity

Solution

Red shift is the increase in wavelength of light observed due to Doppler Effect of light, observed when objects recede from each other.
The red shift as observed in distant galaxies suggests that the galaxies are moving away from us and from each other. This tells that the universe is expanding.
Question 3
1 / -0

Complete the following statement. When a light source is quickly approaching you, you may perceive a:

A
blue shift

B
orange shift

C
red shift

D
violet shift

E
green shift

Solution

Let the original frequency of light source be $$\nu$$.
Using Doppler effect in case when the source of light moves towards the stationary observer :
Apparent frequency seen by the observer $$\nu' = \nu \bigg[ \dfrac{c}{c - v_{source}} \bigg]$$
$$\implies$$ $$\nu'>\nu$$
Thus the observer will perceive a blue shift.
Question 4
1 / -0

Newton postulated his corpuscular theory of light on the basis of

A
newton's rings

B
rectilinear propagation of light

C
colour through thin films

D
dispersion of white light into colours

Solution

Rectilinear propagation of light.
According to Newton’s Corpuscular Theory of Light, light consists of very tiny particles called corpuscles. These corpuscles, on emission from the source of light, travel in straight line with high velocity.
Question 5
1 / -0

Light of wavelength $$600 \eta m$$ is incident normally on a slit of width $$0.2 mm$$. The angular width of central maxima in the diffraction pattern is :
(measured from minimum to minimum)

A
$$6 \times {10}^{-3} rad$$

B
$$4 \times {10}^{-3} rad$$

C
$$2.4 \times {10}^{-3} rad$$

D
$$4.5 \times {10}^{-3} rad$$

Solution

Given : $$\lambda = 600nm = 6\times 10^{-7}$$ $$m$$ $$b = 0.2mm = 0.2\times 10^{-3}$$
Angular width of first minima from central maxima in diffraction pattern $$\theta = \dfrac{\lambda}{d}$$
$$\therefore$$ Angular width of central maxima, $$\theta_c =2\theta = \dfrac{2\lambda}{d}$$
$$\implies$$ $$\theta_c = \dfrac{2\times 6\times 10^{-7}}{0.2\times 10^{-3}} =6\times 10^{-3} $$ $$rad$$
Question 6
1 / -0

The optical instrument which is used in every cricket match is.:

A
Simple microscope

B
Compound microscope

C
Astronomical telescope

D
Binocular

Solution

Binocular is used in every cricket match for the purpose of zooming.
Because, binoculars are a pair of mirror symmetrical telescopes that allow a user to view distant objects using both eyes. To allow both eyes to view distant objects symmetrically, the binoculars require two separate telescopes, one for each eye, held together in the device called binoculars allowing binocular vision. The telescopes are mounted symmetrically side-by-side and aligned to point accurately in the same direction, providing the user with undistorted vision of distant objects. Unlike a monocular (telescope) which only makes use of one telescope to view objects, binoculars are able to provide the user with three-dimensional viewing of distant objects, whilst promoting visual clarity and acuity.
Question 7
1 / -0

The wavefront is a surface in which

A
all points are in the same phase

B
there is a pair of points in opposite phase

C
there is a pair of points with phase difference $$(\dfrac{\pi}{2})$$

D
there is no relation between the phases

Solution

A wavefront is the locus of points characterized by propagation of position of the same phase: a propagation of a line in 1D, a curve in 2D or a surface for a wave in 3D.
Question 8
1 / -0

Diameter of the objective of a telescope is 200 cm. What is the resolving power of a telescope? Take wavelength of light = 5000 $$\mathring{A}$$.

A
$$6.56 \times 10^6$$

B
$$3.28 \times 10^5$$

C
$$1 \times 10^6$$

D
$$3.28 \times 10^6$$

Solution

Given : $$D = 200$$ cm $$=2$$ m $$\lambda = 5000$$ $$oA = 5\times 10^{-7}$$ m
Resolving power of telescope $$RP = \dfrac{D}{1.22 \lambda}$$
$$\therefore$$ $$RP = \dfrac{2}{1.22 \times 5\times 10^{-7}} = 3.28\times 10^6$$
Question 9
1 / -0

With the help of a telescope that has an objective of diameter $$200\ cm$$, it is proved that light of wavelengths of the order of $$6400\overset {\circ}A$$ coming from a star can be easily resolved. Then the limit of resolution is

A
$$39\times 10^{-8}deg$$

B
$$39\times 10^{-8} rad$$

C
$$19.5\times 10^{-8}rad$$

D
$$19.5\times 10^{-8}deg$$

Solution

Limit of resolution $$=\theta = \dfrac{1.22\times \text{wavelength of light}}{\text{diameter of objective}}=\dfrac {1.22\lambda}{d} = \dfrac {1.22\times 6400\times 10^{-10}}{200\times 10^{-12}} = 39\times 10^{-8}rad$$
Question 10
1 / -0

White light reflected from a soap film(Refractive Index$$=1.5$$) has a maxima at $$600$$nm and a minima at $$450$$nm with no minimum in between. Then the thickness of the film is _____$$\times 10^{-7}$$m.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Thin film interference in a soap bubble.
For maxima , constructive interference.
$$2\mu t \cos r =\dfrac{(2m+1)\lambda_1}{2}$$
For minima , constructive interference.
$$2\mu t \cos r =(m+1)\lambda_2$$
$$\lambda_1$$=600 nm ; $$\lambda_2$$=450 nm
t is film thickness and $$\mu$$ is refractive index of film and is 1.5.
Dividing equation 1 by equation 2
$$\dfrac{(2m+1)\lambda_1}{2}=(m+1)\lambda_2$$
Putting the values of $$\lambda_1$$ , $$\lambda_2$$
$$\dfrac{(2m+1)600}{2}=(m+1)450$$
We get $$m=1$$.
Putting the value of $$m$$ in any one of the equations ,
We get for $$r=0$$,
$$ t=\dfrac{(m+1)450}{2\mu}$$
$$t=\dfrac{(1+1)450}{2\times 1.5}$$
$$t=3\times 10^{-7} m$$
Option C is correct.

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Wave Optics Test - 39

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Wave Optics Test - 39

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SHARING IS CARING

Question 1

$$0^o$$

$$30^o$$

$$45^o$$

$$60^o$$

$$90^o$$

Solution

Question 2

Expansion of the universe

The Uncertainty Principle

Black holes

Dark matter

Superconductivity

Solution

Question 3

blue shift

orange shift

red shift

violet shift

green shift

Solution

Question 4

newton's rings

rectilinear propagation of light

colour through thin films

dispersion of white light into colours

Solution

Question 5

$$6 \times {10}^{-3} rad$$

$$4 \times {10}^{-3} rad$$

$$2.4 \times {10}^{-3} rad$$

$$4.5 \times {10}^{-3} rad$$

Solution

Question 6

Simple microscope

Compound microscope

Astronomical telescope

Binocular

Solution

Question 7

all points are in the same phase

there is a pair of points in opposite phase

there is a pair of points with phase difference $$(\dfrac{\pi}{2})$$

there is no relation between the phases

Solution

Question 8

$$6.56 \times 10^6$$

$$3.28 \times 10^5$$

$$1 \times 10^6$$

$$3.28 \times 10^6$$

Solution

Question 9

$$39\times 10^{-8}deg$$

$$39\times 10^{-8} rad$$

$$19.5\times 10^{-8}rad$$

$$19.5\times 10^{-8}deg$$

Solution

Question 10

$$1$$

$$2$$

$$3$$

$$4$$

Solution

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