2 Marks Questions

Question 512 Marks

Among the following waves which can be polarised? Sound waves, radio waves, X-rays, cathode rays.

Answer

Only transverse waves can be polarised. Out of the given waves, radio waves and X-rays are electromagnetic and hence transverse. So, radio waves and X-rays can be polarised.

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Question 522 Marks

Whether the diffraction effects from a slit will be more clearly visible or less clearly, if the slit-width is increased?

Answer

The width of the central band is inversely proportional to the slit Width. So, as the width of the slit is increased, the central band will become less wider and further bands will start merging in them. Hence, diffraction effects will be visible less clearly.

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Question 532 Marks

The wavelength of light in a medium is $\lambda=\frac{\lambda_0}{\mu},$ where $\lambda$ is the wavelength in vacuum. A beam of red light $(\lambda_0=720\text{nm})$ enters into water. The wavelength in water is $\lambda=\frac{\lambda_0}{\mu}=540\text{nm}.$ To a person under water does this light appear green?

Answer

Colour of light will depend only on the frequency of light and not on the wavelength of the light. So, light will appear red to an observer under water.

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Question 542 Marks

A soap film of thickness 0.0011mm appears dark when seen by the reflected light of wavelength 580nm. What is the index of refraction of the soap solution, if it is known to be between 1.2 and 1.5?

Answer

Given, $\text{d}=0.0011\times10^{-3}\text{m}$
For minimum reflection of light, $2\mu\text{d}=\text{n}\lambda$
$\Rightarrow\mu=\frac{\text{n}\lambda}{2\text{d}}=\frac{2\text{n}\lambda}{4\text{d}}=\frac{580\times10^{-9}\times2\text{n}}{4\times11\times10^{-7}}$ $=\frac{5.8}{44}(2\text{n})=0.132(2\text{n})$
Given that, $\mu$ has a value in between 1.2 and 1.5.
⇒ When, $\text{n}=5,\mu=0.132\times10=1.32.$

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Question 552 Marks

Light of wavelength 560nm goes through a pinhole of diameter 0.20mm and falls on a wall at a distance of 2.00m. What will be the radius of the central bright spot formed on the wall?

Answer

$\lambda=560\text{nm}=560\times10^{-9}\text{m},$ $\text{b}=0.20\text{mm}=2\times10^{-4}\text{m},\text{D}=2\text{m}$
Since, $\text{R}=1.22\frac{\lambda\text{D}}{\text{b}}=1.22\times\frac{560\times10^{-9}\times2}{2\times10^{-4}}$ $=6.832\times10^{-3}\text{M}=0.683\text{cm}.$
So, diameter $=2\text{R}=1.37\text{cm}.$

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Question 562 Marks

A long narrow horizontal slit is placed $1\ mm$ above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen $1.0m$ away from the slit. Find the fringe$-$width if the light used has a wavelength of $700\ nm.$

Answer

Given that$, D = 1m, \lambda=700\ \text{nm}=700\times10^{-9}\text{m}$
Since$, a = 2\ mm, d = 2a = 2\ mm = 2 \times 10^{-3}m (L$ loyd’s mirror experiment$)$
Fringe width $=\frac{\lambda\text{D}}{\text{d}}$
$=\frac{700\times10^{-9}\text{m}\times1\text{m}}{2\times10^{-3}\text{m}}$
$=0.35\ \text{mm}.$

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Question 572 Marks

Two transparent slabs having equal thickness but different refractive indices $\mu_1$ and $\mu_2$ are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young's experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point $P_0$ which is equidistant from the slits?

Answer

The change in path difference due to the two slabs is $(\mu_1-\mu_2) ($as in problem no. $16).$
For having a minimum at $P_0$, the path difference should change by $\frac{\lambda}{2}.$
So, $\Rightarrow\frac{\lambda}{2}=(\mu_1-\mu_2)\text{t}\Rightarrow\text{t}=\frac{\lambda}{2(\mu_1-\mu_2)}.$

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Question 582 Marks

Can two identical and independent sodium lamps act as coherent sources? Give reason for your answer.

Answer

No, two independent sources of light cannot act as coherent sources.
Reason: The emission of light is due to millions of atoms. So phase difference between waves emitted from these atoms will change randomly and sustained interference will not be obtained.

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Question 592 Marks

A Young's double slit apparatus has slits separated by $0.28\ mm$ and a screen 4$8\ cm$ away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light $(\lambda=700\ nm$ in vacuum$)$. Find the fringe$-$width of the pattern formed on the screen.

Answer

Given that, $d = 0.28\ mm = 0.28 \times 10^{-3}m, D = 48\ cm = 0.48m, \lambda_\text{a}=700\ nm$ in vacuum
Let, $\lambda_\text{w}$ = wavelength of red light in water
Since, the fringe width of the pattern is given by,
$\beta=\frac{\lambda_\text{w}\text{D}}{\text{d}}$
$=\frac{525\times10^{-9}\times0.48}{0.28\times10^{-3}}=9\times10^{-4}\text{m}=0.90\ mm$

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Question 602 Marks

Does the polarising angle for any transparent medium depend on the wavelength of light?

Answer

Yes, Reason: Angle of polarisation $\text{i}_\text{B}=\tan^{-1}(\text{n})$ and since refractive index varies as inverse of square of wavelength so angle of polarisation decreases with increase of wavelength.

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Question 612 Marks

If we put a cardboard (say 20cm × 20cm) between a light source and our eyes, we can't see the light. But when we put the same cardboard between a sound source and our ear, we hear the sound almost clearly. Explain.

Answer

Light waves have the property of travelling in a straight line, unlike sound waves. When we put a cardboard between the light source and our eyes, the light waves are obstructed by the cardboard and cannot reach our eyes, which doesn't happen when the cardboard is inserted between sound source and our ear.

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Question 622 Marks

Can we conclude from the interference phenomenon whether light is a transverse wave or a longitudinal wave?

Answer

The interference pattern can be produced by any two coherent waves moving in the same direction. It cannot be concluded from the interference phenomenon that light is a transverse wave, as sound waves that are longitudinal in nature also interfere.

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Question 632 Marks

A plate of thickness t made of a material of refractive index $\mu$ is placed in front of one of the slits in a double slit experiment.

Find the change in the optical path due to introduction of the plate.
What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero? Wavelength of the light used is $\lambda$. Neglect any absorption of light in the plate.

Answer

Change in the optical path $=\mu\text{t}-\text{t}=(\mu-1)\text{t}$
To have a dark fringe at the centre the pattern should shift by one half of a fringe.

$\Rightarrow(\mu-1)\text{t}=\frac{\lambda}{2}\Rightarrow\text{t}=\frac{\lambda}{2(\mu-1)}$

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Question 642 Marks

A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Show that if the incident beam makes an angle $\theta=\sin^{-1}\Big(\frac{\lambda}{2\text{D}}\Big)$ with the normal to the plane of the slits, there will be a dark fringe at the centre $P_0 $ of the pattern.

Answer

It can be seen from the figure that the wavefronts reaching $O$ from $S_1$ and $S_2$ will have a path difference of $S_2X.$
In the $\Delta\text{S}_1\text{S}_2\text{X},$
$\sin\theta=\frac{\text{S}_2\text{X}}{\text{S}_1\text{S}_2}$
So, path difference $=\text{S}_2\text{X}=\text{S}_1\text{S}_2\sin\theta=\text{d}\sin\theta=\text{d}\times\frac{\lambda}{2\text{d}}=\frac{\lambda}{2}$
As the path difference is an odd multiple of $\frac{\lambda}{2},$ there will be a dark fringe at point $P_0.$

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Question 652 Marks

In a Young's double -slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Answer

Using $\lambda=\frac{y \cdot d}{n \cdot D}$, where $y=1.2 \times 10^{-2} m$, $d=0.28 \times$$10^{-3} m, D=1.4 m$, and $n=4$. The wavelength $\lambda$ is 6000 Å.

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Question 662 Marks

Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.

Answer

When a third polaroid is rotated between two crossed polaroids, the intensity varies as $I=\left(I_0 / 8\right) \sin ^2(2 \theta)$,eaching maximum at $45^{\circ}$and minimum (zero) at $0^{\circ}$ or $90^{\circ}$.

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Physics 2 Marks Questions