Wave Optics Test

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Wave Optics Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The shape of wave front at a very large distance from source is ______

A
Circular

B
Spherical

C
Cylindrical

D
Plane

Solution

Due to the large distance, the radius of the wavefront can be considered as large (infinity) and hence, a wavefront is almost plane.
Question 2
1 / -0

Which of the following is wrong for interference fringes?

A
Fringes are due to limited portion of wave front

B
All bright fringes are equally bright

C
Distance between two consecutive fringes is constant

D
Fringes are due to the use of coherent sources

Solution

Statements B,C and D are correct whereas the statement "Fringes are due to limited portion of the wave front" is incorrect.
Interference fringes are formed due to the whole portion of wave front.
Question 3
1 / -0

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observe? (speed of light=$$3\times { 10 }^{ 8 }\ m{s}^{ -1 }$$

A
(A) 15.3GHz

B
(B) 10.1GHz

C
(C) 12.1 GHz

D
(D) 17.3 GHz

Solution

$$f '=\left( \sqrt { \frac { 1+\beta }{ 1-\beta } } \right) f,\quad where\quad \beta =\frac { v }{ c } $$
Hence $$\beta = \dfrac{1}{2}$$
and $$f' = \sqrt{3}\times f$$
which implies $$f' = 17.3\ GHz$$
Question 4
1 / -0

Sound waves in air cannot be polarized because :

A
their speed is small

B
they require medium

C
these are longitudinal

D
their speed is temperature dependent

Solution

Sound waves are longitudinal waves.
Question 5
1 / -0

The angular spread of central maximum, in the diffraction pattern, does not depend on ______

A
the distance between the slit and sources

B
width of slit

C
wavelength of light

D
frequency of light

Solution

Angular spread of central maxima$$=(\theta)=\cfrac{\lambda}{b}$$
Where $$\lambda=$$ wavelength of light
$$b=$$ width of slit
$$\Rightarrow$$ Clearly, it does depends on the distance between slit and sources.
Question 6
1 / -0

Which of the following is correct for light diverging from a point source ?

A
The intensity decreases in proportion for the distance squared.

B
The wavefront is parabolic.

C
The intensity at the wavelength does not depend on the distance.

D
None of these.

Solution

A point source produces a spherical wavefront that moves in all the directions from the point.
Whereas the intensity at the wavelength is dependent on the distance and it follows the inverse square law. So, the intensity decreases with an increase in the distance from the source.

Thus, option $$(A)$$ is correct.
Question 7
1 / -0

The diameter of objective of a telescope is $$1m$$. Its resolving limit for the light of wavelength $$4538\overset {\circ}{A}$$, will be :

A
$$5.54\times 10^{-7}rad$$

B
$$2.54\times 10^{-4}rad$$

C
$$6.54\times 10^{-7}rad$$

D
None of the above

Solution

Resolving limit
$$d\theta = \dfrac {1.22\lambda}{a}$$
$$= \dfrac {1.22\times 4538\times 10^{-10}}{1}$$
$$= 5.54\times 10^{-7}rad$$.
Question 8
1 / -0

A double slit is illuminated by light of wave length 6000$$A ^ { 0 }$$.The slits are 0.1 cm apart and the screen is placed 1 m away. Then the angular position of 10th maxima is

A
$$6 \times 10 ^ { - 3 } r a d$$

B
6 rad

C
$$0.006 ^ { \circ }$$

D
$$6 ^ { 0 }$$

Solution

$$\begin{array}{l} \lambda =6000{ { A }^{ o } } \\ d=0.1cm \\ D=1 \\ angular=\frac { { 10\beta } }{ D } \\ =\frac { { 10\times \lambda D } }{ { dD } } =\frac { { 10\lambda } }{ d } =\frac { { 10\times 6000\times { { 10 }^{ -10 } }\times 100 } }{ { o.1 } } \\ =6\times { 10^{ -3 } } \\ =6rad \\ Hence, \\ option\, \, B\, \, is\, correct\, \, answer. \end{array}$$
Question 9
1 / -0

If a star emitting yellow light is accelerated towards earth, then to an observer on earth it will appear?

A
Becoming orange

B
Shining yellow

C
Gradually changing to blue

D
Gradually changing to red

Solution

As the star coming closer to the earth, the frequency of the light increase and the wavelength decrease due to doppler effect. So the colour should gradually change to blue.
Question 10
1 / -0

The wavelength of light visible to eye is of the order of

A
$$10^{-2}\ m$$

B
$$10^{-10}\ m$$

C
$$1\ m$$

D
$$6\times 10^{-7}\ m$$

Solution

The range for visible light is between $$4 \times 10^{-7} m$$ to $$7\times 10^{-7} m$$. The only wavelength that is present in this range given as an option is $$6\times 10^{-7}m$$.
Hence, the correct answer is option (D).

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Wave Optics Test - 23

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Wave Optics Test - 23

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SHARING IS CARING

Question 1

Circular

Spherical

Cylindrical

Plane

Solution

Question 2

Fringes are due to limited portion of wave front

All bright fringes are equally bright

Distance between two consecutive fringes is constant

Fringes are due to the use of coherent sources

Solution

Question 3

(A) 15.3GHz

(B) 10.1GHz

(C) 12.1 GHz

(D) 17.3 GHz

Solution

Question 4

their speed is small

they require medium

these are longitudinal

their speed is temperature dependent

Solution

Question 5

the distance between the slit and sources

width of slit

wavelength of light

frequency of light

Solution

Question 6

The intensity decreases in proportion for the distance squared.

The wavefront is parabolic.

The intensity at the wavelength does not depend on the distance.

None of these.

Solution

Question 7

$$5.54\times 10^{-7}rad$$

$$2.54\times 10^{-4}rad$$

$$6.54\times 10^{-7}rad$$

None of the above

Solution

Question 8

$$6 \times 10 ^ { - 3 } r a d$$

6 rad

$$0.006 ^ { \circ }$$

$$6 ^ { 0 }$$

Solution

Question 9

Becoming orange

Shining yellow

Gradually changing to blue

Gradually changing to red

Solution

Question 10

$$10^{-2}\ m$$

$$10^{-10}\ m$$

$$1\ m$$

$$6\times 10^{-7}\ m$$

Solution

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