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Physics 2 Marks Questions

Wave Optics · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 64 questions.

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Question 12 Marks
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Answer
Given,
Distance of the screen from the slit, D = 1 m
Distance of the first minimum when the centre of screen, x = 2.5 mm = 2.5 × 10-3 m
n = 1 is the first minimum.
Wavelength of light, $\lambda$ = 500 nm = 500 × 10-9 m
= 5 × 10-7 m
Using formula, $\text{x}=\text{n}\frac{\lambda\text{D}}{\text{d}}$
$\text{we have},\ \text{d}=\text{n}\frac{\lambda\text{D}}{\text{x}}$
$\Rightarrow\ \text{d}=\frac{1\times5\times10^{-7}\times1}{2.5\times10^{-3}}$
= 2 × 10-4 m
= 0.2 mm
is the required width of the slit.
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Question 22 Marks
Answer the following question:
As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
Answer
Superposition principle is in accordance with the linear character of the differential equation governing wavemotion. If, y1 and y2 are solutions of the wave equation, then any linear combination of y1 and y2 is also a solution of the waveequation. When amplitudes are large (e.g., high intensity laser beams) and non-linear effects are important, the situation is for more complicated.
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Question 32 Marks
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Answer
Refractive index of glass, $\mu=1.5$
Brewster angle = $\theta$
Brewster angle is related to refractive index as:
$\text{tan}\theta=\mu$
$\theta=\text{tan}^{-1}(1.5)=56.31^\circ$
Therefore, the Brewster angle for air to glass transition is 56.31°.
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Question 42 Marks
The 6563 Å $\text{H}\alpha$ line emitted by hydrogen in a star is found to be redshifted by 15 Å. Estimate the speed with which the star is receding from the Earth.
Answer
Given,
Wavelength of $\text{H}\alpha$ line, $\lambda$ = 15 Å = 15 × 10-10 m
Redshift, $\lambda'-\lambda$ = 15 Å = 15 × 10-10 m
Now, according to the formula,
$\lambda'-\lambda=\frac{\text{v}_\text{s}\lambda}{\text{c}}$
we have, $\text{V}_\text{s}=\frac{\text{C}}{\lambda}\big(\lambda'-\lambda\big)$
$=\frac{3.0\times10^8}{6563\times10^{-10}}\times15\times10^{-10}$
= 6.8566 × 105 ms-1
i.e., Vs = 6.86 × 105 ms-1.
Therefore, the start is receding from the earth at a speed of 6.86 × 105 m/s.
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Question 52 Marks
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Answer
Fresnel's distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation,
$\text{Z}_\text{F}=\frac{\text{a}^2}{\lambda}$
Where,
Aperture width, a = 4 mm = 4 × 10-3 m
Wavelength of light, $\lambda$ = 400 nm = 400 × 10-9 m
$\text{Z}_\text{F}=\frac{\big(4\times10^{-3}\big)^2}{400\times10^{-9}}=40\text{m}$
Therefore, the distance for which the ray optics is a good approximation is 40 m.
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Question 62 Marks
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of $\text{n}\lambda/\text{a}.$ Justify this by suitably dividing the slit to bring out the cancellation.
Answer
Let the slit width 'a' be divided into n equal parts of width 'a' so that,
$\text{a}'=\frac{\text{a}}{\text{n}}$
i.e., a = na'
Then,
$\text{Angle},\ \theta=\frac{\text{n}\lambda}{\text{a}}=\frac{\text{n}\lambda}{\text{na}'}$
$\text{i.e.},\ \theta=\frac{\lambda}{\text{a}'}$
At this derived angle, each slit will make first diffraction minimum. Hence, the resultant intensity for all slits will be zero at an angle of $\frac{\text{n}\lambda}{\text{a}}.$
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Question 72 Marks
Answer the following question:
In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
Answer
Width of central diffraction band $=2\text{D}\frac{\lambda}{\text{d}}.$
Where, d is the width of the slit.
So, on doubling the width of the slit, the size of the central diffraction band reduces to half value. But, the light amplitude becomes double, which increases the intensity four fold.
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Question 82 Marks
In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.
Answer
Distance of the screen from the slits, D = 1 m
Wavelength of light used, $\lambda_1$ = 600 nm
Angular width of the fringe in are, $\theta_1=0.2^{\circ}$
Angular width of the fringe in water $=\theta_2$
Refractive index of water, $\mu=\frac{4}{3}$
Refractive index is related to angular width as:
$\mu=\frac{{\theta}_1}{\theta_2}$
$\theta_2=\frac{3}{4}\theta_1$
$=\frac{3}{4}\times0.2=0.15$
Therefore, the angular width of the fringe in water will reduce to 0.15°.
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Question 92 Marks
In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1º. What is the spacing between the two slits?
Answer
Wavelength of light, $\lambda$ = 600 nm = 600 × 10-9 m
Angular fringe width, $\text{B}_\theta=\frac{\lambda}{\text{d}}$
$\Rightarrow\ \ \text{d}=\frac{\lambda}{\beta}_\theta$
Where, d is the distance between the slits.
Now, 
$\beta_\theta=0.1^\circ=\frac{0.1\times\pi}{180}\ \text{radian}$
$=\frac{0.1\times3.14}{180}\ \text{radian}$
$\therefore\ \text{d}=\frac{600\times10^{-9}\times180}{0.1\times3.14}\ \text{m}=3.44\times10^{-4}\text{m}.$
Is the required spacing between the two slits.
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Question 102 Marks
Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
Answer
Distance between the two towers = 40 km
Size of aperture, a = 50 m
Distance of aperture from tower, Zf
Fresnel distance is the half of the distance between the towers.
$\text{i.e},\ \text{Z}_\text{f}=\frac{40}{2}=20\text{km}=20\times10^3\ \text{m}.$
Therefore, now using the formula,
Fresnel distance, $\text{Z}_\text{F}=\frac{\text{a}^2}{\lambda}$
$\text{we have}\ \lambda=\frac{\text{a}^2}{\text{Z}_\text{F}}=\frac{(50)^2}{20\times10^3}$
$\lambda=125\times10^{-3}\ \text{m}=12.5\ \text{cm}.$
This is the required longest wavelength of radio waves, which can be sent in between the towers without considerable diffraction effects.
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Question 112 Marks
Answer the following question:
Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
Answer
Typical sizes of the apertures involved in ordinary optical instruments are much large than the wavelength of light. Henceforth, the diffraction effects of light waves are negligibly small in these instruments and is of negligible or no significance. Thus, the assumption that light travels in straight lines can be safely used in the optical instruments.
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Question 122 Marks
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?
Answer
Let $I_0$ be the intensity of polarised light after passing through the first polariser $P_1$. Then the intensity of light after passing through second polariser $P_2$ will be
$
I=I_0 \cos ^2 \theta \text {, }
$
where $\theta$ is the angle between pass axes of $P_1$ and $P_2$. Since $P_1$ and $P_3$ are crossed the angle between the pass axes of $P_2$ and $P_3$ will be $(\pi / 2-\theta)$. Hence the intensity of light emerging from $P_3$ will be
$
\begin{aligned}
I & =I_0 \cos ^2 \theta \cos ^2\left(\frac{\pi}{2}-\theta\right) \\
& =I_0 \cos ^2 \theta \sin ^2 \theta=\left(I_0 / 4\right) \sin ^2 2 \theta
\end{aligned}
$
Therefore, the transmitted intensity will be maximum when $\theta=\pi / 4$.
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Question 132 Marks
(a) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why?
(b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave?
(c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light.
Answer
(a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.
(b) No. Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.
(c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing an unit area per unit time.
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Question 142 Marks
  1. When a wave is propagating from a rarer to a denser medium, which characteristic of the wave does not change and why?
  2. What is the ratio of the velocity of the wave in the two media of refractive indices $\mu_{1}$ and $\mu_{2}$?
Answer
  1. Frequency does not change, as frequency is a characteristic of the source of waves.

Alternate Answer

$\frac{\text{v}_{1}}{\lambda}_{1} = \frac{\text{v}_{2}}{\lambda}_{2}=\text{n}$

  1. The ratio of velocities of wave in two media of refractive indices$\mu_{1}$ and $\mu_{2}$ is $\frac{\mu_{2}}{\mu_{1}}$.

Alternate Answer

$\frac{\text{v}{1}}{\text{v}_{2}} = \frac{\mu_{2}}{\mu}_{1}$.

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Question 152 Marks
Draw the intensity pattern for single slit diffraction and double slit interference. Hence, state two differences between interference and diffraction patterns.
Answer

Interference Diffraction
All maxima have equal intensity. Maxima have different (/rapidly decreasing) intensity.
All fringes have equal width. Different (/changing) width.
Superposition of two wavefronts. Superposition of wavelets from the same wavefront.
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Question 162 Marks
Unpolarised light is passed through a polaroid P1. When this polarised beam passes through another polaroid P2 and if the pass axis of P2 makes angle $\theta$ with the pass axis of P1, then write the expression for the polarised beam passing through P2. Draw a plot showing the variation of intensity when $\theta$ varies from 0 to $2\pi$.
Answer

Intensity is $\frac{\text{I}}{2}\text{ }\cos^2\theta$ (if I0 is the intensity of unpolarised light.)

Intensity is ${\text{I}}\text{ }\cos^2\theta$ (if I is the intensity of polarized light.)

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Question 172 Marks
The short wavelength limit for the Lyman series of the hydrogen spectrum is 913·4 $\mathring{\text{A}}$ Calculate the short wavelength limit for Balmer series of the hydrogen spectrum.
Answer
$\frac{1}{\lambda}=R\bigg(\frac{1}{n_1^{\text{ }\text{ }2}}-\frac{1}{n_2^{\text{ }\text{ }2}}\bigg)$
$\therefore $ For Balmer Series: $(\lambda_B)_{\text{short}|}=4/\text{R}$
and For Lyman Series: $(\lambda_L)_{\text{short}|}=1/\text{R}$
$\therefore {\lambda_{B}}=913.4\times4\mathring{\text{A}}=3653.6\mathring{\text{A}}$
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Question 182 Marks
A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit.
Answer

Hint: From condition of diffraction,

$\sin\theta = \text{n}\lambda\text{ (for minima) }$

$ = \bigg(\text{n} + \frac{1}{2}\bigg)\lambda\text{ for maxima}$

Provided n =1, 2, 3...

and n =0 for central maxima,

From condition of minima
$\text{a}\sin\theta = \lambda$ (n =1)

Since the value of $\lambda$ is of nm, so

$\text{a},\theta = \lambda$

$\text{a}.\frac{\text{y}}{\text{D}} = \lambda$

$\bigg[\text{angle} = \frac{\text{arc}}{\text{radius}}\bigg]$

$\therefore\text{a} = \frac{\lambda\text{D}}{\text{y}}$

$\text{a} =\frac{500\times10^{-9}\times1}{2.5\times10^{-3}}\text{m}$

=2x10–4 m.

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Question 192 Marks
Define the term ‘linearly polarised light’.
When’ does the intensity of transmitted light become maximum, when a polaroid sheet is rotated between two crossed polaroids?
Answer
When the vibrations of the electric field vector $(\overrightarrow{\text{E}})$ are confined to only one direction then light is called linearly polarized light.

Intensity of transmitted light is maximum when the Polaroid sheet makes an angle of 450 with the pass axis.

Alternate Answer

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Question 202 Marks
State one feature by which the phenomenon of interference can be distinguished from that of diffraction.
A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘a’. If the distance between the slit and the screen is 0.8 m and the distance of 2nd order maximum from the centre of the screen is 15 mm, calculate the width of the slit.
Answer
The given features:-
Interference Diffraction
  1. Fringes of equal width
 Fringes of unequal width
  1. Fringes of equal intensity
 Fringes of decreasing intensity
  1. Due to superposition of two waves (wavefronts)
 Due to superposition of secondary wavefronts.
  1. Maximal at $\theta =\frac{\lambda}{\text{a}}$
 Minima at $\theta = \frac{\lambda}{\text{a}}$
$\text{ (Numerical : Formula}\text{ y} =\frac{(2\text{n} + 1)\lambda\text{D}}{2\text{a}}\text{ or } \frac{5\lambda\text{D}}{2\text{a}}$
Substitution and Result: 8 × 10–5 m or 0.08 mm.
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Question 212 Marks
Find the intensity at a point on a screen in Young’s double slit experiment where the interfering waves have a path difference of (i) $\lambda/6$, and (ii) $\lambda/2.$
Answer
Phase difference $=\frac{2\pi}{\lambda}\times$ Path diffrence
Path difference $\frac{\lambda}{6}$ ⟹ phase difference $=\frac{\pi}{3}$
Path difference $\frac{\lambda}{2}$ ⟹ phase difference $={\pi}$
$\text{I}=4\text{I}_0\cos^{2}\bigg(\frac{\phi}{2}\bigg)$
  1. $\text{I}_1=4\text{I}_0\times\frac{3}{4}=3\text{I}_0$
  2. $\text{I}_2=4\text{I}_0\times0=0$
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Question 222 Marks
Find the intensity at a point on a screen in Young’s double slit experiment where the interfering waves of equal intensity have a path difference of (i) $\lambda/4$, and (ii) $\lambda/3$.
Answer
Path difference 𝜆/4 ⟹ phase difference 𝜋/2
Path difference 𝜆/3 ⟹ phase difference (2𝜋/3)
$I=4I_0\cos^2\Big(\frac{\theta}{2}\Big)$
  1. $I_1=4I_0\times\frac{1}{2}=2I_0$
  2. $I_2=4I_0\times\frac{1}{4}=I_0$
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Question 232 Marks
Distinguish between polarised and unpolarized light. Does the intensity of polarised light emitted by a Polaroid depend on its orientation? Explain briefly.
The vibrations in a beam of polarised light make an angle of 60° with the axis of the polaroid sheet. What percentage of light is transmitted through the sheet?
Answer
If the direction of vibration of electric field vector/plane of vibration of electric field vector, does not change with time, the light is polarised.

Whereas, if the direction of vibration of electric field vector/plane of vibration of electric field vector changes randomly in very short intervals of time/with time, the light is unpolarised.

Yes, it depends upon orientation of Polaroid because electric field vibrations, that are not in the direction of pass axis of Polaroid, are absorbed. Hence, intensity changes.

$I=I_0\cos^2\theta$

= angle between vibrations in light and axis of polaroid sheet)

$I=I_0\cos^260^\circ=\frac{I_0}{4}$

$\Longrightarrow\frac{I}{I_0}\times100=\frac{1}{4}\times100=25\text{%}$

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Question 242 Marks
When are two objects just resolved? Explain. How can the resolving power of a compound microscope be increased? Use relevant formula to support your answer.
Answer
Two objects are said to be just resolved when, in their diffraction patterns, central maxima of one object coincides with the first minima, of the diffraction pattern of the second object. Limit of resolution of compound microscope.
$ \text{d}_{min}=\frac{1.22{\lambda}}{2\text{ }n\text{ }sin\text{ }{\beta}}$
Resolving power is the reciprocal of limit of resolution $( \text{d}_{min})$
Therefore, to increase resolving power $\lambda$ can be reduced and refractive index of the medium can be increased.
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Question 252 Marks
State Brewster’s law.
The value of Brewster angle for a transparent medium is different for light of different colours. Give reason.
Answer
When unpolarized light is incident on the surface separating two media, the reflected light gets (completely) polarised only when the reflected light and refracted light become perpendicular to each other.

The refractive index of denser medium, with respect to rarer medium, is given by $\mu = \tan\text{i}_{p}$

Since Refractive index (μ) of a transparent medium is different for different colours, hence Brewster angle is different for different colours.

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Question 262 Marks
Draw the output waveform at X, using the given inputs A and B for the logic circuit shown below. Also, identify the logic operation performed by this circuit.

Answer
Drawing correct output waveform.

OR gate.

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Question 272 Marks
Define the term ‘Half-life’ of a radioactive substance. Two different radioactive substances have half-lives T1 and T2 and number of undecayed atoms at an instant N1 and N2, respectively. Find the ratio of their activities at that instant.
Answer
Half-life: It is the time interval after which the activity of a radioactive sample reduces to half of initial value.
$\because\text{R}=\lambda\text{N}$
$\Rightarrow\frac{\text{R}_1}{\text{R}_2}=\frac{\lambda_1}{\lambda_2}\times\frac{\text{N}_1}{\text{N}_2}$
$\because\lambda=\frac{0.693}{\text{T}_\frac{1}{2}}$
$\Rightarrow\frac{\lambda_1}{\lambda_2}=\frac{\text{T}_2}{\text{T}_1}$
$\therefore\frac{\text{R}_1}{\text{R}_2}=\frac{\text{T}_2}{\text{T}_1}\times\frac{\text{N}_1}{\text{N}_2}$
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Question 282 Marks
Define wavefront of a travelling wave. Using Huygens principle, obtain the law of refraction at a plane interface when light passes from a denser to rarer medium.
Answer

A plane wavefront AB is incident in rarer medium at instant t = 0 on interface XY separating it from a denser medium. When wavelet A is on interface, B is at a distance BB, from it. It takes t time to cover the distance BB1 = v1t = to reach on interface XY. Mean while, the wavelet from A reaches to point A1 covering a distance AA1 = v2t in denser medium.
To locate A1, draw a secondary wavelet with radius AA1 = v2t & centre A. Draw tangent from B, into this sec. wavelet intersecting at A1.
A1B1 is refracted wavefront at instant t.
i = angle of incidence,
r = angle of refraction.
$\therefore\triangle\text{ABB}_1$
$\Rightarrow\sin\text{i}=\frac{\text{BB}_1}{\text{AB}_1}$
$\triangle\text{AA}_1\text{B}_1$
$\Rightarrow\sin\text{r}=\frac{\text{AA}_1}{\text{AB}_1}$
$\therefore\frac{\sin\text{i}}{\sin\text{r}}=\frac{\text{BB}_1}{\text{AA}_1}=\frac{\text{v}_1\text{t}}{\text{v}_2\text{t}}=\frac{\text{v}_1}{\text{v}_2}$
$\therefore\frac{\sin\text{i}}{\sin\text{r}}=\frac{\text{v}_1}{\text{v}_2}=\text{constant}$
Also, $\text{n}_1=\frac{\text{c}}{\text{v}_1}\text{n}_2=\frac{\text{c}}{\text{v}_2}$
$\frac{\text{n}_1}{\text{n}_2}=\frac{\text{v}_2}{\text{v}_1}$
$\Rightarrow\frac{\sin\text{i}}{\frac{\sin\text{r}}{}}=\frac{\text{n}_2}{\text{n}_1}=\text{constant}$
Which is Snell's law.
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Question 292 Marks
An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index $\frac{4\sqrt2}{5}.$
Answer
When the prism is kept in another medium. we have to take the refractive index of the prism with respect to the given medium:
$^{\text{medium}}\mu_\text{prism}=\frac{\mu_\text{prism}}{\mu_\text{medium}}=\frac{\sin\Big[\Big(\frac{\text{A}+\text{D}_\text{m}}{2}\Big)\Big]}{\sin\Big(\frac{\text{A}}{2}\Big)}$
$\frac{1.6}{\frac{4\sqrt2}{5}}=\frac{\sin\Big[\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)\Big]}{\sin\Big(\frac{60^\circ}{2}\Big)}$
$\sqrt2=\frac{\sin\Big[\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)\Big]}{\frac{1}{2}}$
$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)$
$90^\circ=60^\circ+\text{D}_\text{m}$
$\text{D}_\text{m}=30^\circ$
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Question 302 Marks
Calculate the radius of curvature of an equi-concave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of -5D?
Answer
Using lens maker formulae for given equi-cancave lens,
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_2}{\text{n}_1}-1\Big)\Big(\frac{1}{-\text{R}}-\frac{1}{\text{R}}\Big)$
$\Rightarrow\frac{1}{\text{f}}=\Big(\frac{\text{n}_2}{\text{n}_1}-1\Big)\Big(-\frac{2}{\text{R}}\Big)\ ...(1)$
Where n2 and n1 n2 and n1 are the refractive index of the lens and respectively. where n2 = 1.5 & n1 = 1.4
Power of lens = -5D (Given)
Focal length, $\text{f}=\frac{1}{-5}\times100=-20\text{cm}$
Putting all the values in equation 1, we get,
$\Rightarrow\frac{1}{-20}=\Big(\frac{1.5}{1.4}-1\Big)\Big(-\frac{2}{\text{R}}\Big)$
$\Rightarrow\frac{1}{20}=\Big(\frac{0.1}{1.4}\Big)\Big(\frac{2}{\text{R}}\Big)$
$\Rightarrow\text{R}=\frac{4}{1.4}=2.86\text{cm}$
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Question 312 Marks
The value of the Brewster angle for a transparent medium is different for lights of different colours.
Answer
Brewster’s angle, $\text{i}_\text{p}=\tan^{-1}(\text{n})$
As refractive index n varies as inverse value of wavelength; it is different for lights of different wavelengths (colours), therefore, Brewster’s angle is different for lights of different colours.
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Question 322 Marks
White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe $(\lambda=400\text{nm}),$ which overlaps with a red fringe $(\lambda=700\text{nm}).$
Answer
Let mth bright fringe of violet light overlaps with nth bright fringe of red light.
$\therefore\frac{\text{m}\times400\text{nm}\times\text{D}}{\text{d}}=\frac{\text{n}\times700\text{nm}\times\text{D}}{\text{d}}\Rightarrow\frac{\text{m}}{\text{n}}=\frac{7}{4}$
⇒ 7th bright fringe of violet light overlaps with 4th bright fringe of red light (minimum). Also, it can be seen that 14th violet fringe will overlap 8th red fringe.
Because, $\frac{\text{m}}{\text{n}}=\frac{7}{4}=\frac{14}{8}$
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Question 332 Marks
Is it necessary to have two waves of equal intensity to study interference pattern? Will there be an effect on clarity if the waves have unequal intensity?
Answer
Interference pattern can be studied with waves of unequal intensity.
$\text{I}=\text{I}_1+\text{I}_2+2\sqrt{[\text{I}_1\times\text{I}_2]}\cos(\phi),$
Where $\phi$, = phase difference.
In case of waves of unequal intensities, the contrast will not be clear, as the minima will not be completely dark.
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Question 342 Marks
No interference pattern is detected when two coherent sources are infinitely close to each other. Why?
Answer
Fringe width of interference fringes is given by $\beta=\frac{\text{D}\lambda}{\text{d}}\alpha\frac{1}{\text{d}}.$ When d is infinitely small, fringe width β will be too large. In such a case even a single fringe may occupy the whole field of view. Hence, the interference pattern cannot be detected.
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Question 352 Marks
The linewidth of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the linewidth of a bright fringe in a Young's double slit experiment in terms of $\lambda$, d and D where the symbols have their usual meanings.
Answer
The line width of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum.
We know that, for intensity to be half the maximum,
$\text{y}=\pm\frac{\lambda}{4\text{d}}$
$\therefore$ Line width $=\frac{\lambda\text{D}}{4\text{d}}+\frac{\lambda\text{D}}{4\text{d}}=\frac{\lambda\text{D}}{2\text{d}}.$
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Question 362 Marks
A parallel beam of light of wavelength 560nm falls on a thin film of oil (refractive index = 1.4). What should be the minimum thickness of the film so that it strongly reflects the light?
Answer
Given that, $\lambda=560\times10^{-9}\text{m},\mu=1.4.$
For strong reflection, $2\mu\text{d}=(2\text{n}+1)\frac{\lambda}{2}\Rightarrow\text{d}=\frac{(2\text{n}+1)\lambda}{4\text{d}}$
For minimum thickness, putting n = 0.
$\Rightarrow\text{d}=\frac{\lambda}{4\text{d}}\Rightarrow\text{d}=\frac{560\times10^{-9}}{14}=10^{-7}\text{m}=100\text{nm}.$
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Question 372 Marks
The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0mm. The screen is placed at a distance of 2.5m from the slits and the separation between the slits is 1.0mm. Calculate the wavelength of light used for the experiment.
Answer
Given that, $\beta=1\text{mm}=10^{-3}\text{m},$ D = 2.5m and d = 1mm = 10-3m
So, $10^{-3}\text{m}=\frac{25\times\lambda}{10^{-3}}\Rightarrow\lambda=4\times10^{-7}\text{m}=400\text{nm}.$
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Question 382 Marks
Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength 500nm. The separation between the slits is 2.0 × 10-3m.
Answer
Given that, $\lambda=500\text{nm}=500\times10^{-9}\text{m}$ and d = 2 × 10-3m

As shown in the figure, angular separation $\theta=\frac{\beta}{\text{D}}=\frac{\lambda\text{D}}{\text{dD}}=\frac{\lambda}{\text{d}}$
So, $\theta=\frac{\beta}{\text{D}}=\frac{\lambda}{\text{d}}=\frac{500\times10^{-9}}{2\times10^{-3}}=250\times10^{-6}$
= 25 × 10-5 radian = 0.014 degree.
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Question 392 Marks
In a double slit interference experiment, the separation between the slits is 1.0mm, the wavelength of light used is 5.0 × 10-7m and the distance of the screen from the slits is 1.0m.
  1. Find the distance of the centre of the first minimum from the centre of the central maximum.
  2. How many bright fringes are formed in one centimeter width on the screen?
Answer
Given that, d = 1mm = 10-3m, D = 1m.
So, fringe with $=\frac{\text{D}\lambda}{\text{d}}=0.5\text{mm}.$
  1. So, distance of centre of first minimum from centre of central maximum $=\frac{0.5}{2}\text{mm}=0.25\text{mm}$
  2. No. of fringes $\frac{10}{0.5}=20.$
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Question 402 Marks
Why is the interference pattern not detected, when two coherent sources are far apart?
Answer
Fringe width of interference fringes, is given by $\beta=\frac{\text{D}\lambda}{\text{d}}\alpha\frac{1}{\text{d}},$.If the sources are far apart; d is large; so fringe width $(\beta)$ will be so small that the fringes are not resolved and they do not appear separate. That is why the interference pattern is not detected for large separation of coherent sources.
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Question 412 Marks
A transparent paper (refractive index = 1.45) of thickness 0.02mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620nm. How many fringes will cross through the centre if the paper is removed?
Answer
Given that, $\mu=1.45,$ t = 0.02mm = 0.02 × 10-3m and $\lambda=620\text{nm}=620\times10^{-9}\text{m}$
We know, when the transparent paper is pasted in one of the slits, the optical path changes by $(\mu-1)\text{t}.$
Again, for shift of one fringe, the optical path should be changed by $\lambda.$
So, no. of fringes crossing through the centre is given by,
$\text{n}=\frac{(\mu-1)\text{t}}{\lambda}=\frac{0.45\times0.02\times10^{-3}}{620\times10^{-9}}=14.5$
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Question 422 Marks
A source emitting light of wavelengths 480nm and 600nm is used in. a double slit interference experiment. The separation between the slits is 0.25mm and the interference is observed on a screen placed at 150cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
Answer
We know that, the first maximum (next to central maximum) occurs at $\text{y}=\frac{\lambda\text{D}}{\text{d}}$
Given that, $\lambda_1=480\text{nm},\lambda_2=600\text{nm},$ D = 150cm = 1.5m and d = 0.25mm = 0.25 × 10-3m
So, $\text{y}_1=\frac{\text{D}\lambda_1}{\text{d}}=\frac{1.5\times480\times10^{-9}}{0.25\times10^{-3}}=2.88\text{mm}$
$\text{y}_2=\frac{1.5\times600\times10^{-9}}{0.25\times10^{-3}}=3.6\text{mm}.$
So, the separation between these two bright fringes is given by,
$\therefore$ separation = y2 - y1 = 3.60 - 2.88 = 0.72mm.
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Question 432 Marks
A parallel beam of monochromatic light falls normally on a single narrow slit. How does the angular width of the principal maximum in the resulting diffraction pattern depend on the width of the slit?
Answer
Width of central maximum, $\beta=\frac{2\lambda\text{D}}{\text{a}}\alpha \frac{1}{\text{a}},$ where a is width of the slit. That is angular width of principal maximum decreases with increase of width of the slit.
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Question 442 Marks
The refractive index of a material is. What is the angle of refraction if the unpolarised light is incident on it at the polarising angle of the medium?
Answer
From Brewster's law,
polarising angle $\text{i}_\text{B}=\tan^{-1}\text{(n)}=\tan^{-1}\sqrt3=60^\circ$
Also $\text{i}_\text{B}+\text{r}=90^\circ$
$\therefore$ Angle of refrection $=90^\circ-\text{i}_\text{B}=90^\circ-60^\circ=30^\circ.$
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Question 452 Marks
In a Young's double slit experiment, two narrow vertical slits placed 0.800mm apart are illuminated by the same source of yellow light of wavelength 589nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00m away?
Answer
Given that, $\text{d} = 0.8\text{mm} = 0.8 × 10^{-3}\text{m},$ $\lambda=589\text{nm}=589\times10^{-9}\text{m}\text{ and D}=2\text{m}.$ 
So, $\beta=\frac{\text{D}\lambda}{\text{d}}=\frac{589\times10^{-9}\times2}{0.8\times10^{-3}}=1.47\times10^{-3}=147\text{mm}.$
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Question 462 Marks
State the importance of coherent sources in the phenomenon of interference.
Answer
If coherent sources are not taken, the phase difference between two interfering waves, meeting at any point will change continuously and a sustained interference pattern will not be obtained. Thus, coherent sources provide sustained interference pattern.
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Question 472 Marks
A convex lens of diameter 8.0cm is used to focus a parallel beam of light of wavelength 620nm. If the light be focused at a distance of 20cm from the lens, what would be the radius of the central bright spot formed?
Answer
$\lambda=620\text{nm}=620\times10^{-9}\text{m},$
$\text{D}=20\text{cm}=20\times10^{-2}\text{m},\text{b}=8\text{cm}=8\times10^{-2}\text{m}$
$\therefore\text{R}=1.22\times\frac{620\times10^{-4}\times20\times10^{-2}}{8\times10^{-2}}$ $=1891\times10^{-9}=1.9\times10^{-6}\text{m}$
So, diameter $=2\text{R}=3.8\times10^{-6}\text{m}$
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Question 482 Marks
In a Young's double slit experiment $\lambda=500\text{nm},$ d = 1.0mm and D = 1.0m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
Answer
Given that, D = 1m, d = 1mm = 10-3m, $\lambda=500\text{nm}=5\times10^{-7}\text{m}$
For intensity to be half the maximum intensity.
$\text{y}=\frac{\lambda\text{D}}{4\text{d}}$
$\Rightarrow\text{y}=\frac{5\times10^{-7}\times1}{4\times10^{-3}}\Rightarrow\text{y}=1.25\times10^{-4}\text{m}.$
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Question 492 Marks
A double slit S1 - S2 is illuminated by a coherent light of wavelength $\lambda$. The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen $\sum$ is placed behind the double slit at a distance D2 from it (figure 17-E2). The screen E receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.

Answer
It can be seen from the figure that, the apparent distance of the screen from the slits is,
D = 2D1 + D2
So, Fringe width $=\frac{\text{D}\lambda}{\text{d}}=\frac{(2\text{D}_1+\text{D}_2)\lambda}{\text{d}}$
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Question 502 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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2 Marks Questions - Physics STD 12 Science Questions - Vidyadip