Physics Wave Optics

In Young's double slit experiment, the fringe width is 12mm. If the entire arrangement is placed in water refractive index $\frac{4}{3}$ , then the fringe width becomes (in mm)

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Payal Gupta

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$β = 12 m m = 12 * 1 0^{- 3} m$

$\begin{array}{l} β' = \frac{β}{μ} = \frac{12 * 1 0^{- 3}}{4 / 3} = 9 * 1 0^{- 3} m \\ β' = 9 m m \end{array}$

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An unpolarised light beam strikes a glass surface at Brewster's angle. Then

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The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x : 4 where x is_______.

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Amp. a slit with

Intensity a (Amp)² a (slit width)²

$\frac{l_{1}}{l_{2}} = {(\frac{3}{1})}^{2} = \frac{9}{1} \Rightarrow l_{1} = 9 I_{2}$

$\frac{l_{m i n}}{l_{m a x}} = \frac{{(\sqrt[]{l_{1}} - \sqrt[]{l_{2}})}^{2}}{{(\sqrt[]{l_{1}} + \sqrt[]{l_{2}})}^{2}} = \frac{{(3 - 1)}^{2}}{{(3 + 1)}^{2}}$

$\Rightarrow \frac{1}{4} = \frac{x}{4}$

x = 1.00

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In a Young's double slit experiment, the slits are separated by 0.3mm and the screen is 1.5m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is 2.4cm. The frequency of light used is______* 10¹⁴ Hz.

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Distance of 4^th bright fringe from central maxima = $\frac{4 λ D}{d}$

$S o, \frac{2.4}{100} = \frac{8 λ D}{d} \Rightarrow λ = \frac{2.4 * 0.3 * 1 0^{- 3}}{100 * 8 * 1.5} = 6 * 1 0^{- 7} m$

$f = \frac{c}{λ} = \frac{3 * 1 0^{8}}{6 * 1 0^{- 7}} = 5 * 1 0^{14} H z$

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The light waves from two coherent sources have same intensity I₁ = I₂ = I₀. In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima?

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Kindly go through the solution

$l_{m a x} = {(\sqrt[]{l_{1}} + \sqrt[]{l_{2}})}^{2} = 4 l_{0}$

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White light is passed through a double slit and interference is observed on a screen 1.5m away. The separation between the slits is 0.3mm. The first violet and red fringes are formed 2.0mm and 3.5mm away from the central white fringes. The difference in wavelengths of red and violet light is………nm.

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For first fringe

$\frac{d y_{1}}{D} = 1 * λ_{1} [n = 1] λ_{1} \to$ Red wavelength

$\frac{d y_{2}}{D} = 1 * λ_{2} [n = 1] λ_{2} \to$ violet wavelength

$\frac{d}{D} (y_{1} - y_{2}) = λ_{1} - λ_{2}$

$y_{1} - y_{2} = \frac{D}{d} (λ_{1} - λ_{2})$

= $\frac{0.3 * 1 0^{- 3}}{1.5} (3.5 - 2) * 1 0^{- 3}$

= 300 * 10^-9

= 300 nm

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A source of light is placed in front of a screen. Intensity of light on the screen in I. Two Polaroids P₁ and P₂ are so placed in between the source of light and screen that the intensity of light on screen in $\frac{I}{2} .$ P₂ should be rotated by an angle of_______ (degrees) so that the intensity of light on the screen becomes

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$\frac{l}{2} c o s^{2} ? = \frac{3 l}{8}$

$? c o s ? = \frac{\sqrt[]{3}}{2}$

$? ? = 30 °$

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A source and a detector move away from each other in absence of wind with a speed of 20m/s with respect to the ground. If the detector detects a frequency of 1800 Hz of the sound coming from the source, then the original frequency of source considering speed of sound in air 340 m/s will be ……Hz.

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$f' = \frac{340 - 20}{340 + 20} * f_{0}$

$\Rightarrow 1800 = \frac{320}{360} * f_{0}$

f₀ = 2025 Hz

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The two light beams having intensities I and 9l interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point P and p at point Q. Then the difference between the resultant intensities at P and Q will be:

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Given, l₁ = 1

l₂ = 9l

case 1 $θ = \frac{π}{2}$

$l_{P} = l_{1} + l_{2} + 2 \sqrt[]{l_{1} l_{2}} c o s θ$

= 10 l

case 2 θ = π

$l_{Q} = l_{1} + l_{2} + 2 \sqrt[]{l_{1} l_{2}} c o s θ$

$= 10 l - 6 l$

$l_{P} - I_{Q} = 10 l - 4 l = 6 l$

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A light whose electric field vectors are completely removed by using a good Polaroid, allowed to incident on the surface of the prism at Brewster's angle. Choose the most suitable option for phenomenon related to the prism.

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