Wave Optics Test

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Wave Optics Test - 29

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Question 1
1 / -0

Directions For Questions

The resolving power of a microscope is its capacity to form distinct images of two very close objects. This power depends upon the wavelength of the light used for illuminating the object and is more for violet light than for the red.
The breakthrough in high-resolution microscopy was made with the advent of quantum mechanics which established the dual nature of wave and matter. Ordinary light, under suitable conditions, could behave like a stream of particles, whereas, with the help of particles moving with high speeds all wave phenomena like diffraction, interference, etc. Could be demonstrated. The relationship between the momentum of the particle and the wavelength of the associated wave is given by the de Broglie's hypothesis, which states that the wavelength is inversely proportional to momentum.

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A microscope is used with sodium light and its resolving power is not sufficiently large.Higher resolution will be obtained by using wavelength of

A
20 micron

B
2 micron

C
1 micron

D
400 A$$^{\circ}$$

Solution

Since power of resolution is more for violet than for red, we conclude that resolving power is greater for light with lower wavelengths.
Wavelength of sodium light is around 589nm
Hence resolving power increases for light with lower wavelength.
Hence option D is correct.
Question 2
1 / -0

The oil layer on the surface of water appears coloured, due to interference . For this effect to be visible the thickness of oil layers will be

A
$$1mm$$

B
$$1cm$$

C
$$100A^o$$

D
$$1000A^o$$

Solution

A phase difference of $$\pi$$ occurs after reflection from the oil surface. But for the second beam, this does not happen when reflected from water surface,
This gives the condition for constructive interference as-
$$2\mu_{oil}d=\dfrac{\lambda_{air}}{2}$$
Hence, $$\lambda=4\mu_{oil}d$$
For $$400nm<\lambda <700nm$$
$$400nm<4\mu_{oil}d<700nm$$
$$\mu_{oil}=1.45$$
Hence $$689.65A^{\circ}<d<1206.8A^{\circ}$$
Question 3
1 / -0

White light is incident normally on a soap film of thickness $$15\times 10^{-5}cm$$ and refractive index $$1.33$$. Which wavelength is reflected maximum in the visible region?

A
$$26000A^o$$

B
$$8866A^o$$

C
$$5320A^o$$

D
$$3800A^o$$

Solution

$$n_{1}=n_{3}<n_{2}$$
A phase change of $$\pi$$ takes place at point A, but not at B. Hence for a constructive interference to be obtained, the condition is-
$$2\mu d \cos r=(m-\dfrac{1}{2})\lambda$$
Since incidence is normal, $$\cos r=1$$
$$\lambda=\dfrac{2\mu d}{(m-\dfrac{1}{2})}$$
$$\lambda$$ lies in visible region only for m=8.
$$\lambda_{m=8}=5320A^{\circ}$$
Question 4
1 / -0

Light of wavelength $$7500A^o$$ is incidents on a thin glass plate $$(\mu=1.5)$$ so that the angle of refraction obtained is $$30^0$$. If the plate appears the dark then the minimum thickness of the plate will be

A
$$4000 \sqrt3A^o$$

B
$$\displaystyle\frac{8000}{\sqrt3}A^o$$

C
$$\displaystyle\frac{5000}{\sqrt3}A^o$$

D
$$1000 \sqrt3A^o$$

Solution

Path difference between ray 1 and ray 2 is $$2 \mu dsin(r)$$
For plate to appear dark , the rays must interfere destructively. Hence the path difference should be an even multiple of $$\lambda$$.
This is because there is an extra phase change of $$\pi$$ due to reflection of ray 1 from the above layer(reflection from a higher refractive index medium, which is not the case with ray 2).
Therefore, $$2\mu d \sin r=n\lambda$$
$$ d= \dfrac{n \lambda}{2 \mu sin(r)}$$
For minimum thickness, order(n)=1
Hence, $$d=\dfrac{5000}{\sqrt{3}} A^{\circ}$$
Question 5
1 / -0

The slits in a Young's double- slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is $$I_o$$. If one of the slit is closed, the intensity at this point will be_______

A
$$I_o$$

B
$$\displaystyle \frac{I_o}{4}$$

C
$$\displaystyle \frac{I_o}{2}$$

D
$$4I_o$$

Solution

for interference of two waves,
$${ I }_{ total }={ I }_{ 1 }+{ I }_{ 2 }+2\sqrt { { I }_{ 1 }{ I }_{ 2 } } \cos { \theta }$$
for equal intensities and central fringe,
$${ I }_{ total }={ 4I }_{ 1 }={ I }_{ 0 }$$ ..... given.
$$\therefore { \quad I }_{ 1 }=\dfrac { { I }_{ 0 } }{ 4 } $$
Hence when one slit is closed the intensity will be $$\dfrac { { I }_{ 0 } }{ 4 } $$
Question 6
1 / -0

The transverse nature of light waves is verified by

A
reflection of light

B
polarisation of light

C
refraction of light

D
interference of light

Solution

Phenomenon of polarisation helps in establishing the fact that light waves are transverse in nature, otherwise it was believed that they are longitudinal in nature like sound waves.
Question 7
1 / -0

All particles of a wave front vibrate

A
in same phase

B
in opposite phase

C
up and down

D
left and right

Solution

Wave front by definition is the locus of points having same phase.

Answer. A) in same phase
Question 8
1 / -0

Light waves travel in a vacuum, along the $$X-$$axis. Which of the following may represent the wave fronts?

A
$$x=c$$

B
$$y=c$$

C
$$z=c$$

D
$$x+y+z=c$$

Solution

Given the direction of propagation is $$ \hat{i} $$
Wave fronts will be planes $$ \perp \hat{i} $$
In the given options, only plane $$ x=c $$ is $$ \perp \hat{i} $$
Question 9
1 / -0

A student is asked to measure the wavelength of monochromatic light . He sets up the apparatus as shown. $$S_1,S_2,S_3$$ are narrow parallel slits. $$L$$ is radiant lamps and $$M$$ is a micrometer eyepiece. The student fails to observe interference fringes. We would advice him to

A
increase the width of $$S_1$$

B
decrease the distance between $$S_2$$ and $$S_3$$.

C
replace $$L$$ with a white light lamp.

D
make $$S_2$$ and $$S_3$$ wider.

Solution

The fringe width of a double slit interference pattern is given by
$$\beta=\dfrac{D\lambda}{d}$$
where $$D$$ is the distance between screen and the plane containing slits
$$d$$ is distance between the slits $$S_2$$ and $$S_3$$.
To make the fringe pattern visible, the fringe width must be increased, for which $$d$$ should be decreased.
Question 10
1 / -0

In Young's double slit experiment, the interference pattern obtained with white light will be

A
the central fringe bright and alternate bright and dark fringes

B
the central fringe achromatic and coloured fringes for small path difference

C
the central fringe dark

D
the central fringe coloured

Solution

If white light is used a white centre fringe is observed, but all the other fringes have coloured edges, the blue edge being nearer the centre. Eventually the fringes overlap and a uniform white light is produced.
Hence correct answer is option B.

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Wave Optics Test - 29

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Wave Optics Test - 29

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Question 1

20 micron

2 micron

1 micron

400 A$$^{\circ}$$

Solution

Question 2

$$1mm$$

$$1cm$$

$$100A^o$$

$$1000A^o$$

Solution

Question 3

$$26000A^o$$

$$8866A^o$$

$$5320A^o$$

$$3800A^o$$

Solution

Question 4

$$4000 \sqrt3A^o$$

$$\displaystyle\frac{8000}{\sqrt3}A^o$$

$$\displaystyle\frac{5000}{\sqrt3}A^o$$

$$1000 \sqrt3A^o$$

Solution

Question 5

$$I_o$$

$$\displaystyle \frac{I_o}{4}$$

$$\displaystyle \frac{I_o}{2}$$

$$4I_o$$

Solution

Question 6

reflection of light

polarisation of light

refraction of light

interference of light

Solution

Question 7

in same phase

in opposite phase

up and down

left and right

Solution

Question 8

$$x=c$$

$$y=c$$

$$z=c$$

$$x+y+z=c$$

Solution

Question 9

increase the width of $$S_1$$

decrease the distance between $$S_2$$ and $$S_3$$.

replace $$L$$ with a white light lamp.

make $$S_2$$ and $$S_3$$ wider.

Solution

Question 10

the central fringe bright and alternate bright and dark fringes

the central fringe achromatic and coloured fringes for small path difference

the central fringe dark

the central fringe coloured

Solution

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