Wave Optics Test

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Wave Optics Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

A rocket is going away from the earth at a speed of $$10\ m/s$$ If the wavelength of the light wave emitted by it be $$5700\ overset{o}{A}$$, what will be its Doppler's shift

A
$$200\ \overset{o}{A}$$

B
$$19\ \overset{o}{A}$$

C
$$20\ \overset{o}{A}$$

D
$$0.2\ \overset{o}{A}$$

Solution

$$\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}$$
$$\Rightarrow \Delta\lambda =\dfrac{5700\times 10^{6}}{3\times 10^{3}}$$
$$=19\mathring{A}$$
Question 2
1 / -0

If a star is moving towards the earth, then the lines are shifted towards

A
Red

B
Infrared

C
Blue

D
Green

Solution

When the source and observer approach each other, apparent frequency increases and hence wavelength decreases.
Question 3
1 / -0

Diffraction and interference of light suggest

A
nature of light is electromagnetic

B
Wave nature

C
Nature is quantum

D
Nature of light is transverse

Solution
Question 4
1 / -0

If the separation between slit in Young's double slit experiment is reduced to $$\dfrac 13 rd$$, the fringe width becomes $$n$$ times. The value of $$n$$ is

A
$$3$$

B
$$\dfrac 13$$

C
$$9$$

D
$$\dfrac 19$$

Solution

$$\beta \propto \dfrac {\lambda}{d}$$
as $$d\to \dfrac {d}{3}$$
so $$\beta \to 3\beta $$
$$\therefore n=3$$
Question 5
1 / -0

How does the red shift confirm that the universe is expanding

A
Due to Wien's law

B
Due to Stefan's law

C
Due to Kirchhoff's law

D
Due to Doppler's law

Solution

If the light received from galaxies indicates a shift towards the red end of spectrum of light, it means that the galaxies should be receding away (Doppler's effect). Therefore we conclude that the universe is expanding.
Question 6
1 / -0

A star is going away from the earth. An observer on the earth will see the wavelength of light coming from the star

A
Decreased

B
Increased

C
Neither decreased nor increased

D
Decreased or increased depending upon the velocity of the star

Solution
Question 7
1 / -0

The $$k$$ line of singly ionised calcium has a wavelength of $$393.3\ nm$$ as measured on earth. In the spectrum of one of the observed galaxies, this spectral line is located at $$401.8\ nm$$. The speed with which the galaxy is moving away from us, will be

A
$$6480\ km/s$$

B
$$3240\ km/s$$

C
$$4240\ km/sec$$

D
$$None\ of\ these$$

Solution

$$\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}\Rightarrow \dfrac{(401.8-393.3)}{393.3}=\dfrac{v}{3\times 10^{8}}$$
$$\Rightarrow v=6.48\times 10^{6}\ m/s=6480\ km/sec.$$
Question 8
1 / -0

In young double slit experiment the maximum intensity of light is $$ I_{max}$$ then the intensity of light of the path difference $$ \dfrac{\lambda }{2}$$ will be

A
$$ I_{max}$$

B
$$ \dfrac {Imax}{2}$$

C
$$ \dfrac {Imax}{4}$$

D
zero

Solution

Sound waves are longitudinal while light rays are transverse Diffraction effect is more pronounced if the size of obstacle or aperture is of the wavelength of the waves. As the wavelength of light $$(\sim 10^{-6} m)$$ is much smaller than the size of the objects around us, so diffraction of light is not easily seen. But sound waves large wavelength. They get easily diffracted by the objects around us.
Question 9
1 / -0

The galaxies are moving away from each other. It is explained by

A
White dwarf star

B
Red shift

C
Neutron star

D
None of these

Solution

The galaxies are moving away from each other. It is explained by Red shift.
Question 10
1 / -0

Suppose you perform Young's double-slit experiment with the slit separation slightly smaller than the wavelength of the light. As a screen, you use a large half-cylinder with its axis along the midline between the slits. What interference pattern will you see on the interior surface of the cylinder?

A
bright and dark fringes so closely spaced as to be indistinguishable.

B
one central bright fringe and two dark fringes only

C
a completely bright screen with no dark fringes

D
one central dark fringe and two bright fringes only

E
a completely dark screen with no bright fringes

Solution

With two fine slits separated by a distance $$d$$ slightly less than $$\lambda$$,
the equation $$d \;sin \;\theta =0$$ has the usual solution
$$\theta = 0$$, but $$d \;sin \;\theta = \lambda$$ has no solution: there is no first-order maximum.

However,$$d \;sin \;\theta = \dfrac{1}{2}\lambda$$ has solution: first-order minims flank the central maximum on each side.

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Re-Attempt Weekly Quiz Competition

Wave Optics Test - 65

Result

Wave Optics Test - 65

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SHARING IS CARING

Question 1

$$200\ \overset{o}{A}$$

$$19\ \overset{o}{A}$$

$$20\ \overset{o}{A}$$

$$0.2\ \overset{o}{A}$$

Solution

Question 2

Red

Infrared

Blue

Green

Solution

Question 3

nature of light is electromagnetic

Wave nature

Nature is quantum

Nature of light is transverse

Solution

Question 4

$$3$$

$$\dfrac 13$$

$$9$$

$$\dfrac 19$$

Solution

Question 5

Due to Wien's law

Due to Stefan's law

Due to Kirchhoff's law

Due to Doppler's law

Solution

Question 6

Decreased

Increased

Neither decreased nor increased

Decreased or increased depending upon the velocity of the star

Solution

Question 7

$$6480\ km/s$$

$$3240\ km/s$$

$$4240\ km/sec$$

$$None\ of\ these$$

Solution

Question 8

$$ I_{max}$$

$$ \dfrac {Imax}{2}$$

$$ \dfrac {Imax}{4}$$

zero

Solution

Question 9

White dwarf star

Red shift

Neutron star

None of these

Solution

Question 10

bright and dark fringes so closely spaced as to be indistinguishable.

one central bright fringe and two dark fringes only

a completely bright screen with no dark fringes

one central dark fringe and two bright fringes only

a completely dark screen with no bright fringes

Solution

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