Wave Optics Test

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Wave Optics Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

Fill in the blank.
The phenomenon of polarization shows that light has ________ nature.

A
Particle

B
Transverse

C
Longitudinal

D
Dual

Solution

Transverse nature of light waves, indicates the direction of oscillation (of electromagnetic field) in light is perpendicular to the direction of propagation of light. Polarization necessarily imply that the oscillation can happen with more than one orientation. As there are multiple orientations perpendicular to the direction of propagation, a transverse wave can be polarized.
Question 2
1 / -0

If the two slits in Young's double slit experiment are of unequal width, then

A
The bright fringes will have unequal spacing

B
The bright fringes will have unequal brightness

C
The fringes do not appear

D
The dark fringes are not perfectly dark

Solution

For sustained interference fringes, the two sources of light must be coherent and of same frequency, which is clearly not the case here Hence no fringes appear. Option C.
Question 3
1 / -0

If $$i$$ is the angle of incidence, the angle between the incident wave front and the normal to the reflecting surface is

A
$$i$$

B
$${ 90 }^{ o }-i$$

C
$${ 90 }^{ o }+i$$

D
$$i-{ 90 }^{ o }$$

Solution

A case of wavefront incident on the reflecting surface is shown in the figure above. The line AE represents incident wavefront at time of incidence. The angle between the wavefront and normal to the reflecting surface here is $$90^{\circ}-i$$.
Question 4
1 / -0

The wave theory of Light was proposed by ...............

A
Newton

B
Huygens

C
Frenel

D
Yung

Solution

Newton has given the theory that light is made of tiny particles. In 1678, Dutch physicist, Christiaan Huygens, believed that light was made up of waves vibrating up and down perpendicular to the direction of light, and therefore formulated a way of visualizing wave propagation. This known as 'Huygens' Principle'.
Question 5
1 / -0

Directions For Questions

Select and write the most appropriate answer from the given alternatives for each sub-question :

...view full instructions

The resolving power of a telescope depends upon the ...........

A
Length of the telescope

B
Focal length of an objective

C
Diameter of an objective

D
Focal length of an eyepiece

Solution

R.P. of telescope $$=\dfrac{1}{d\theta } =\dfrac{a}{1.22 \lambda }$$
Thus, it is clear that a telescope with a large diameter of the objective has higher resolving power. Thus, the resolving power of a telescope depends on the diameter of objective.
Question 6
1 / -0

The numerical aperture of objective of a microscope is $$0.12$$. The limit of resolution, when light of wavelength $$6000\mathring { A } $$ is used to view an object is:

A
$$0.25\times { 10 }^{ -7 }m$$

B
$$2.5\times { 10 }^{ -7 }m$$

C
$$25\times { 10 }^{ -7 }m$$

D
$$250\times { 10 }^{ -7 }m$$

Solution

Resolving power of a microscope of objective with numerical aperture $$NA$$ for a light of wavelength $$\lambda$$ is given by:
$$R = \cfrac{\lambda}{2 NA}$$
$$=\cfrac{6000 \times 10^{-10}}{2 \times 0.12}$$
$$= 25 \times 10^{-7}\ m$$
Question 7
1 / -0

The resolving power of telescope of aperture $$100cm$$ for light of wavelength $$5.5\times {10}^{-7}m$$ is ...............

A
$$0.149\times { 10 }^{ 7 }$$

B
$$1.49\times { 10 }^{ 7 }$$

C
$$14.9\times { 10 }^{ 7 }$$

D
$$149\times { 10 }^{ 7 }$$

Solution

Aperture (diameter) of telescope $$D = 100 cm =1 m$$
Wavelength of light $$\lambda = 5.5\times 10^{-7} m$$
Resolving power of telescope $$RP = \dfrac{D}{1.22 \lambda}$$
$$\implies$$ $$RP = \dfrac{1}{1.22\times 5.5\times 10^{-7}} = 0.149 \times 10^{7}$$
Question 8
1 / -0

When two waves of almost equal frequency $$n_1$$ and $$n_2$$ are produced simultaneously, then the times interval between successive maxima is

A
$$\dfrac{1}{n_1+n_2}$$

B
$$\dfrac{1}{n_1}+\dfrac{1}{n_2}$$

C
$$\dfrac{1}{n_1}-\dfrac{1}{n_2}$$

D
$$\dfrac{1}{n_1-n_2}$$

Solution

The intensity of the sound is maximum in all these cases,
$$2 \pi(\dfrac{n_1-n_2}{2})\times t=n\pi$$
$$t=0$$ ,$$\dfrac{1}{n_1-n_2} $$, $$\dfrac{2}{n_1-n_2}....$$
Time interval between two successive maxima = time interval between two successive beats $$= \dfrac{1}{n_1-n_2}$$
Question 9
1 / -0

Newton's ring pattern in reflected system, viewed under white light consists of

A
Equally spaced bright and dark bands with central dark spot

B
Equally spaced bright and dark bands with central white spot

C
A few coloured rings with central dark spot

D
A few coloured rings with central white spot

Solution

Rings are fringes of equal thickness. They are observed when light is reflected from a plano-convex lens of a long focal length placed in contact with a plane glass plate. A thin air film is formed between the plate and the lens. The thickness of the air film varies from zero at the point of contact to some value t. If the lens plate system is illuminated with monochromatic light falling on it normally, concentric bright and dark interference rings are observed in reflected light. These circular fringes were discovered by Newton and are called Newton’s rings. Those rings are formed with equally spaced bright and dark bands with central spot as a dark spot. Hence A is the correct answer.
Question 10
1 / -0

An unpolarised beam of intensity $${ I }_{ 0 }$$ falls on a polaroid at an angle of $$45^0$$. The intensity of the emergent light is

A
$$\dfrac { { I }_{ 0 } }{ 2 } $$

B
$${ I }_{ 0 }$$

C
$$\dfrac { { I }_{ 0 } }{ 4 } $$

D
Zero

Solution

$$I=I_ocos^2\omega t$$
As the angle of incidence is $$45^0$$ so $$cos^245^0= 1/2$$
Hence $$I_{av}=\dfrac{I_o}{2}$$

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Wave Optics Test - 40

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Wave Optics Test - 40

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SHARING IS CARING

Question 1

Particle

Transverse

Longitudinal

Dual

Solution

Question 2

The bright fringes will have unequal spacing

The bright fringes will have unequal brightness

The fringes do not appear

The dark fringes are not perfectly dark

Solution

Question 3

$$i$$

$${ 90 }^{ o }-i$$

$${ 90 }^{ o }+i$$

$$i-{ 90 }^{ o }$$

Solution

Question 4

Newton

Huygens

Frenel

Yung

Solution

Question 5

Length of the telescope

Focal length of an objective

Diameter of an objective

Focal length of an eyepiece

Solution

Question 6

$$0.25\times { 10 }^{ -7 }m$$

$$2.5\times { 10 }^{ -7 }m$$

$$25\times { 10 }^{ -7 }m$$

$$250\times { 10 }^{ -7 }m$$

Solution

Question 7

$$0.149\times { 10 }^{ 7 }$$

$$1.49\times { 10 }^{ 7 }$$

$$14.9\times { 10 }^{ 7 }$$

$$149\times { 10 }^{ 7 }$$

Solution

Question 8

$$\dfrac{1}{n_1+n_2}$$

$$\dfrac{1}{n_1}+\dfrac{1}{n_2}$$

$$\dfrac{1}{n_1}-\dfrac{1}{n_2}$$

$$\dfrac{1}{n_1-n_2}$$

Solution

Question 9

Equally spaced bright and dark bands with central dark spot

Equally spaced bright and dark bands with central white spot

A few coloured rings with central dark spot

A few coloured rings with central white spot

Solution

Question 10

$$\dfrac { { I }_{ 0 } }{ 2 } $$

$${ I }_{ 0 }$$

$$\dfrac { { I }_{ 0 } }{ 4 } $$

Zero

Solution

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