Wave Optics Test

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Wave Optics Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

White light may be considered to be a mixture of waves with $$\lambda$$ ranging between $$3900$$ $$\overset{o}{A}$$ and $$7800$$ $$\overset{o}{A}$$. An oil film thickness $$10,000$$ $$\overset{o}{A}$$ is examined normally by the reflected light. If $$\mu =1.4$$, then the film appears bright for.

A
$$4308$$ $$\overset{o}{A}$$, $$5091$$ $$\overset{o}{A}$$, $$6222\overset{o}{A}$$

B
$$4000$$ $$\overset{o}{A}$$, $$5091$$ $$\overset{o}{A}$$, $$5600$$ $$\overset{o}{A}$$

C
$$4667$$ $$\overset{o}{A}$$, $$6222$$ $$\overset{o}{A}$$, $$7000$$ $$\overset{o}{A}$$

D
$$4000$$ $$\overset{o}{A}$$, $$4667$$ $$\overset{o}{A}$$, $$5600$$ $$\overset{o}{A}$$

Solution

The film appears bright when the path difference
$$(2\mu t\cos r)$$ is equal to odd multiple of $$\dfrac{\lambda}{2}$$
i.e $$2\mu t\cos r=(2n-1)\lambda/2$$ where $$n=1,2,3 ....$$
$$\therefore \lambda=\dfrac{4\mu t\cos r}{(2n-1)}$$
$$=\dfrac{4\times 1.4\times 10,000\times 10^{-10}\times \cos 0}{(2n-1)}=\dfrac{56000}{(2n-1)}\mathring{A}$$
$$\therefore \lambda =56000\mathring{A}, 18666\mathring{A}, 8000\mathring{A}, 6222\mathring{A}, 5091\mathring{A}, 4308\mathring{A}, 3733\mathring{A}$$.
The wavelength which are not within specified range are to be refracted.
Question 2
1 / -0

Direction :
The question has a paragraph followed by two statements, Statement-1 and Statement-2. Of the given four alternatives after the statements, choose the one that describes the statements.
A thin air film is formed by putting the convex surface of a piano-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.
Statement 1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of $$\pi$$.
Statement 2 : The centre of the interference pattern is dark.

A
Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1.

B
Both statements 1 and 2 are true but statement 2 is not the correct explanation of statement 1.

C
Statement 1 is true but statement 2 is false.

D
Both statements 1 and 2 are false.

Solution

When the light is reflected from a denser medium, a phase change of $$\pi$$ occurs in the reflected wave. Hence statement-1 is true.
Since the phase difference between the two reflected waves at the centre of interference pattern is $$\pi$$. Hence the centre of the interference pattern is dark. Hence statement-2 is true.
Question 3
1 / -0

The limit of resolution of an optical instrument is the smallest angle that two points on an object have to subtend at the eye so that they are.

A
Unresolved

B
Well resolved

C
Just resolved

D
None of these

Solution

Limit of resolution of an optical instrument is the minimum angle that two points on an object have to subtend at the eye so that they are just resolved. (C)
Question 4
1 / -0

When a red glass is heated in dark room it will seem.

A
Black

B
Green

C
Yellow

D
Red

Solution
Question 5
1 / -0

The earth is moving towards a fixed star with a velocity of $$30 km s^{-1}$$ . An observer on the earth observes a shift of $$0.58 A^0$$ in the wavelength of light coming from the star. The actual wavelength of light emitted by the star is then

A
$$5800 A^0$$

B
$$2400A^0$$

C
$$12000A^0$$

D
$$6000A^0$$

Solution

Given: The speed of the earth towards the star is $$30\ km/s$$.
The shift in the wavelength observed by the observer is $$0.58\ A^o$$

The actual wavelength of the light emitted by the star is given by:
$$\Delta\lambda=\dfrac{v}{c}[\lambda]$$

Here, $$v=30km s^{-1}=30\times 10^3 m s^{-1}$$
$$c=3\times 10^8 m s^{-1},\Delta \lambda=0.58A^{\circ}$$

$$\lambda=\dfrac{c}{v}.\Delta\lambda=\left(\dfrac{3\times10^8}{30\times 10^3}\right)\times 0.58$$
$$=5800A^o$$
Question 6
1 / -0

Choose the Correct answer from alternative given.
The idea of secondary wavelets for the propagation of a wave was first given by:

A
Newton

B
Huygens

C
Maxwell

D
Fresnel

Solution

According to the Huygens theory, all the points on a wavefront act as source for a secondary wavefront or a wavelet which spread outward from the point and move at the speed of light.
Question 7
1 / -0

In the case of the waves from two coherent sources $$S_1$$ and $$S_2$$ , there will be constructive interference at an arbitrary point P, the path difference $$S_1P - S_2P$$ is then

A
$$[ n + \dfrac{1}{2}] \lambda$$

B
$$ n \lambda$$

C
$$[ n - \dfrac{1}{2}] \lambda$$

D
$$ \dfrac{\lambda}{2}$$

Solution

constructive interference occurs when the path differences $$(S_1P-s_2P)$$ is an integral multiple of $$\lambda$$ or $$(S_1P-s_2P)=n\lambda$$
Where n=0,1 ,2 3,........
Question 8
1 / -0

If numerical aperture of a microscope is increased then its

A
resolving power remains constant

B
resolving power becomes zero

C
limit of resolution is decreased

D
limit of resolution is increased

Solution

On increasing the numerical aperture of the microscope,more light rays are gathered that were diffracted at larger angles and hence its limit of resolution increases.

Hence, answer is option-(D).
Question 9
1 / -0

An electron microscope gives higher magnifications than an optical microscope because.

A
The electrons have more energy than the light particles

B
The velocity of electrons is smaller than that of light

C
The wavelength of electrons used is smaller as compared to the wavelength of visible light

D
The electron microscope uses more powerful lenses

Solution

An electron microscope gives higher magnifications than an optical microscope because the wavelength of electrons used is smaller as compared to the wavelength of visible light.
Question 10
1 / -0

In Young's double slit experiment shows in figure, $$S_1$$ and $$S_2$$ are coherent sources and S is the screen having a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits. Which wavelength passing through the hole has the strongest intensity?

A
400 nm

B
700 nm

C
500 nm

D
667 nm

Solution

From youngs double slit experiment we have
path difference $$(\Delta x)=\dfrac{y\ d}{D}$$
$$y=1\ mm, d=0.5\ mm\ \ \ D=50\ cm =500\ mm$$
So,
$$\Delta x=\dfrac{1\times 0.5}{500}\ mm$$
$$\Delta x=\dfrac{0.5}{500}\times 10^{6}\ nm$$
$$\Delta x=1000\ nm$$
For the strongest intensity at $$H$$ there should be constructive interference
$$\Rightarrow \Delta x=n\lambda$$
where
$$\lambda \Rightarrow$$ wavelength of light
$$n\rightarrow$$ integer $$(1, 2, 3,....)$$
$$1000=n\lambda$$
For $$n=1$$
$$\lambda =1000\ nm$$
For $$n=2$$
$$\lambda=\dfrac{1000}{2}=500\ nm$$
For $$n=3$$
$$\lambda=\dfrac{1000}{3}=333.3\ nm$$
So,
For the visible light $$\lambda =500\ nm$$

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Wave Optics Test - 43

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Wave Optics Test - 43

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SHARING IS CARING

Question 1

$$4308$$ $$\overset{o}{A}$$, $$5091$$ $$\overset{o}{A}$$, $$6222\overset{o}{A}$$

$$4000$$ $$\overset{o}{A}$$, $$5091$$ $$\overset{o}{A}$$, $$5600$$ $$\overset{o}{A}$$

$$4667$$ $$\overset{o}{A}$$, $$6222$$ $$\overset{o}{A}$$, $$7000$$ $$\overset{o}{A}$$

$$4000$$ $$\overset{o}{A}$$, $$4667$$ $$\overset{o}{A}$$, $$5600$$ $$\overset{o}{A}$$

Solution

Question 2

Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1.

Both statements 1 and 2 are true but statement 2 is not the correct explanation of statement 1.

Statement 1 is true but statement 2 is false.

Both statements 1 and 2 are false.

Solution

Question 3

Unresolved

Well resolved

Just resolved

None of these

Solution

Question 4

Black

Green

Yellow

Red

Solution

Question 5

$$5800 A^0$$

$$2400A^0$$

$$12000A^0$$

$$6000A^0$$

Solution

Question 6

Newton

Huygens

Maxwell

Fresnel

Solution

Question 7

$$[ n + \dfrac{1}{2}] \lambda$$

$$ n \lambda$$

$$[ n - \dfrac{1}{2}] \lambda$$

$$ \dfrac{\lambda}{2}$$

Solution

Question 8

resolving power remains constant

resolving power becomes zero

limit of resolution is decreased

limit of resolution is increased

Solution

Question 9

The electrons have more energy than the light particles

The velocity of electrons is smaller than that of light

The wavelength of electrons used is smaller as compared to the wavelength of visible light

The electron microscope uses more powerful lenses

Solution

Question 10

400 nm

700 nm

500 nm

667 nm

Solution

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