Physics 3 Marks Question

1. Distance of the n^th bright fringe on the screen from the central maximum is given by the relation,

$\text{x}=\text{n}\lambda_1\bigg(\frac{\text{D}}{\text{d}}\bigg)$

For third bright fringe, n = 3

$\therefore\ \text{x}=3\times650\frac{\text{D}}{\text{d}}=1950\bigg(\frac{\text{D}}{\text{d}}\bigg)\ \text{nm}$

2. Let the n^th bright fringe due to wavelength, $\lambda_2$ and (n - 1)^th bright fringe due to wavelength $\lambda_1$ coincide on the screen. We can equate the conditions for bright fringes as:

$\text{n}\lambda_2=\big(\text{n}-1\big)\lambda_1$

520n = 650n - 650

650 = 130n

$\therefore\ \text{n}=5$

Hence, the least distance from the central maximum can be obtained by the relation:

$\text{x}=\text{n}\lambda_2\frac{\text{D}}{\text{d}}$

$=5\times520\frac{\text{D}}{\text{d}}=2600\frac{\text{D}}{\text{d}}\text{nm}$

Note: The value of d and Dare not given in the question.

1. Refractive index of glass, µ = 1.5

Speed of light, c = 3 × 10⁸ m/s

Speed of light in glass is given by the relation,

$\text{v}=\frac{\text{c}}{\mu}$

$=\frac{3\times10^8}{1.5}=2\times10^8\ \text{m/s}$

Hence, the speed of light in glass is 2 × 10⁸ m/s.

2. The speed of light in glass is not independent of the colour of light.

The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

1. When the pass axis of a polaroid makes an angle $\theta$ with the plane of polarisation of polarised light of intensity I_o incident on it, then the intensity of the transmitted emergent light is given by $\text{I} = \text{I}_{o}\cos^{2}\theta.$
2. 
3. $\mu = \tan\text{i}_{\beta}$

$ =\tan60^{o} =\sqrt{3} = 1.7.$

1. When unpolarised light is incident on the boundary between two transparent media, the reflected light gets plane polarized with its electric vector perpendicular to the plane of incidence.

The polarization is complete when the reflected and refracted rays are at right angles to each other. This condition occurs for an angle of incidence, i_p, where tan i_p = $\mu$

2. Intensity of light through $\text{P}_1=\frac{I_0}{2}$

Intensity of light through $\text{P}_2=\frac{I_0}{2}\cos^260$

$\frac{I_0}{2}.\bigg(\frac{1}{2}\bigg)^2=\frac{I_0}{8}$

Intensity of light through $ \text{P}_3=\frac{I_0}{8}\cos^230=\frac{I_0}{8}\times\bigg(\frac{\sqrt{3}}{2}\bigg)^2=\frac{3I_0}{32}$

Alternate Answer

The acceleration, of the charges, in the scattering molecules, due to the electric field of the incident radiation, can be in two mutually perpendicular directions.

The observer, however, receives the scattered light, corresponding to only one of these two sets of the accelerated charges.

This causes scattered light to get polarised. Alternatively: The observer receives scattered light corresponding to only one of the two sets of accelerated charges i.e. electrons oscillating perpendicular to the direction of propagation

2. Intensity of light through $\text{P}_1=\frac{I_0}{2}$

Intensity of light through $\text{P}_2=\frac{I_0}{2}\cos^245$

$\frac{I_0}{2}.\frac{1}{2}=\frac{I_0}{4}$

Intensity of light through $\text{P}_3=\frac{I_0}{4}\cos^245=\frac{I_0}{4}\times\frac{1}{2}=\frac{I_0}{8}$

1. Distance of third bright fringe $ - \text{y}_{3} = \frac{\text{n}\lambda\text{D}}{\text{d}}$

$ = \frac{3 \times 520 \times10^{−9}\times 1}{1.5 \times 10^{−3}}$

$ = 1.04 \times 10^{−3}\text{𝑚}\simeq1\text{ 𝑚𝑚}$.

2. Let n^th maxima of 650𝑛𝑚 coincides with the (n + 1)^th maxima of 520 nm.

 $\therefore \text{𝑛} \times 650 \times10^{−9} =(\text{ n} + 1 )520 \times 10^{−9}$

$\Rightarrow\text{n} = 4$

$\therefore$ The least distance of the point is given by

$\text{y} =\frac{\text{nD}\lambda_{1}}{\text{d}}$

$ = \frac{4\times 1 \times 650 \times 10^{−9}}{1.5 \times 10^{−3}}\text{m} = 1.733 \times 10^{−3}\text{𝑚}\simeq 1.7\text{𝑚𝑚}$.

Light from the sodium lamp passing through the single Polaroid sheet ( P₁) does not show any variation in intensity when this sheet is rotated. However, if the light, transmitted by P₁, is made to pass through another Polaroid sheet (P₂) the light intensity, coming out of P₂, varies from a maximum to zero, and again to maximum, when P₂ is rotated. These observations are consistent only with the transverse nature of light waves.

2. Intensity of light transmitted through P₁ = I₀ / 2.

Intensity of light transmitted through P₃ = (I₀ / 2) x cos² 30⁰.

= 3 I₀ / 8

Intensity of light transmitted through $\text{P}_{2} = \frac{3}{8}\text{I}_{o}\cos^{2}60^{o}$

$=\frac{3}{32}\text{I}_{o}.$

$\text{X}_{n} = \frac{\text{n}\lambda\text{D}}{\text{d}}$

Given: n = 2, d = 0.15 mm, λ = 450 nm and D = 1.0 m

$\text{x}_{2} = \frac{2\times450\times10^{-9}\times1.0}{0.15\times10^{-3}} = 6 \times10^{-3}\text{m} = 6\text{mm}$

Distance of n^th minima from central maxima

$\text{y}_{2} = \frac{(2\text{n} - 1 )\lambda\text{D}}{2\text{d}} = \frac{(2\times2-1)450\times10^{-9}\times1}{2\times0.15\times10^{-3}} = 4.5\times10^{-3}\text{m} = 4.5\text{mm}$

When the screen is moved away from the slits fringes become farther apart.
(Fringe width α distance of screen.)

1. $\theta=\frac{n\lambda}{a}$ are points of minima.
2. $\theta=\big(n+\frac{1}{2}\big)\frac{\lambda}{a}$ are points of maxima

1. Angular width of fringes

$\theta =\lambda/\text{d},$  where d = separation between two slits

Here $\theta = 0.1^{\circ} = 0.1\times\frac{\pi}{180}\text{radian}$

$\therefore\text{d} = \frac{600\times10^{-9}\times180}{0.1\times\pi}$

 = 3.43 X 10-4 m

= 0.34 m

2. For Reflected light:  avelength remains same. Frequency remains same.

For Refracted light: Wavelength decreases½ Frequency remains same.

1. Coherent sources are needed to ensure that the positions of maxima and minima do not change with time.

Alternate Answer

Coherent sources have constant phase difference and, therefore, produce a sustained interference pattern.

2. $\text{I} = \text{I}_{1} + \text{I}_{2} + 2 \sqrt{\text{I}_{1}\text{I}_{2}}\cos\theta$

For path difference $\lambda,$ phase difference

$\Phi = 2\pi$

$\text{Hence}, \text{k} = 4\text{I}_{0}\cos^{2}\pi =4 \text{ I}_{0}$

For path difference $\lambda\big/3$

Phase difference $\Phi = 2\pi\big/3$

Intensity

$\text{I}' = 4\text{I}_{0}\cos^{2}\pi\big/3$

$ = 4\text{I}_{0}\bigg(\frac{1}{2}\bigg)^{2} = \text{I}_{0}$

Therefore, $\text{I}' = \frac{\text{k}}{4}.$

1. A surface of constant phase/A locus of points which oscillate in same phase.
2.  

3. $\sin\text{i} = \frac{\text{BC}}{\text{AC}}$

$\sin\text{r} = \frac{\text{AE}}{\text{AC}}$

$\therefore \frac{\sin\text{i}}{\sin\text{r}} = \frac{\text{BC}}{\text{AE}}$

$ = \frac{c_1\text{t}}{c_2\text{t}} = \frac{c_1}{\text{c}_{2}} = n =\text{ constant}$

1. $\text{I}_1=\frac{\text{I}}{2}$

$\text{I}_2=\text{I}$

$\frac{\text{I}_{\text{max}}}{\text{I}_{\text{min}}}=\bigg(\frac{\sqrt{\text{I}_1}+\sqrt{\text{I}_2}}{\sqrt{\text{I}_1}-\sqrt{\text{I}_2}}\bigg)^2$

$=\Bigg(\frac{\sqrt{\frac{\text{I}}{2}}+\sqrt{\text{I}}}{\sqrt{\frac{\text{I}}{2}}-\sqrt{\text{I}}}\Bigg)^2$

$=\bigg(\frac{1+\sqrt{2}}{1-\sqrt{2}}\bigg)^2$

$=\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$

$=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$

2. If we use white light, central fringe will be white, then there will be colourful fringes on both sides, starting from violet to red (fringe width $\propto$ wavelength) and then there will be general illumination of screen.

1. $\text{z}=\sqrt{\text{R}^2+\text{X}^2\text{C}}=\text{impedance}$

$\therefore\text{I}=\frac{\text{V}}{\text{Z}}=\frac{\text{v}_1\sin\omega\text{t}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}$

$\text{V}_\text{R}=\text{IR}$

$\text{V}_\text{C}=\text{IX}_\text{C}$

In phase of 1 lags I by $\frac{\pi}{2}$

$\text{V}=\sqrt{\text{V}^2\text{R}+\text{V}^2_\text{C}}$

$=\sqrt{(\text{IR})^2+(\text{IX}_\text{C})^2}$

$\Rightarrow\text{I}=\frac{\text{V}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}=\frac{\text{r}_0\sin\omega\text{t}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}$

$\phi=$ phase diff bet V & I

$\tan\phi=\frac{\text{V}_\text{C}}{\text{V}_\text{R}}=\frac{\text{X}_\text{C}}{\text{R}}$

1. Linear fringe width increases $\beta=\frac{\lambda\text{D}}{\text{d}}$

$\Rightarrow\beta\propto\text{D}$

No effect on angular fringe width $\Big(\text{Q}=\frac{\lambda}{\text{d}}\Big).$

2. Both linear fringe width & angular fringe width decrease $\Big(\beta\propto\frac{1}{\text{d}},\text{Q}\propto\frac{1}{\text{d}}\Big).$
3. If condition $\frac{\text{s}}{\text{S}}<\frac{\lambda}{\text{d}}$ is satisfied, interference will be obtained otherwise, no interference will be obtained.

1. When diameter of objective lens decreases, θ and hence sinθ decreases; so the resolving power decreases.
2. The focal length of objective lens has no effect on resolving power of microscope.

1. At a point of constructive interference:

$\varphi=2\text{n}\pi(\text{n}=0,1,2,\dots)\Rightarrow\cos\varphi=1$

$\therefore\text{I}_\max=2\text{I}+2\text{I}=4\text{I}$

2. At a point of denstructive interference:

$\Rightarrow\cos\varphi=0 $ 

$\therefore \text { I}_\min=\text{2I}-\text{2I}=0$

1. Separation between two consecutive maxima is equal to fringe width.

So, $\beta=\frac{\lambda\text{D}}{\text{d}}=\frac{5\times10^{-7}\times1}{10^{-2}}\text{m}=5\times10^{-5}\text{m}=0.05\text{mm}.$

2. When, $\lambda=1\text{mm}=10^{-3}\text{m}$

$10^{-3}\text{m}=\frac{5\times10^{-7}\times1}{\text{D}}\Rightarrow\text{D}=5\times10^{-4}\text{m}=0.50\text{mm}.$