Physics Wave Optics

The interference pattern is obtained with two coherent light source of intensity ratio 4:1. And the ratio $\frac{I_{m a x} + I_{m i n}}{I_{m a x} - I_{m i n}} i s \frac{5}{x} .$ Then , the value of x will be equal to:

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Contributor-Level 10

$\frac{l_{1}}{l_{2}} = \frac{4}{1}$

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

$\Rightarrow \frac{l_{m a x} + l_{m i n}}{l_{m a x} - l_{m i n}} = \frac{5}{x}$

$\Rightarrow \frac{1}{2} * \frac{l_{2} (1 + \frac{l_{1}}{l_{2}})}{\sqrt[]{l_{1} l_{2}}} = \frac{5}{x}$

$\Rightarrow \frac{1}{2} \frac{1}{\sqrt[]{4}} = \frac{1}{x} \Rightarrow x = 4$

When electric field vector is completely removed and incident on Brewster's angle then only refraction takes place.

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In Young's double slit experiment the two slits are 0.6mm distance apart. Interference pattern is observed on a screen at a distance 80cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be__________nm.

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3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\Rightarrow \frac{d y}{D} = (2 n + 1) \frac{λ}{2}$ [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{d y}{D} = \frac{λ}{2}$

$λ = \frac{2 d}{D} y$

$= \frac{2 d}{D} * \frac{d}{2} [y = \frac{d}{2}, G i v e n$ first dark fringe is observed on the screen directly opposite to one of the slits]

$λ = \frac{2 * 0.6 * 0.6}{800 * 2}$

$λ = 450 m m$

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An observer moves towards stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:

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$V_{0} = \frac{1}{5} C$

$V_{0} = \frac{C}{5}, V_{S} = 0$

$f = f_{0} [\frac{C + V_{0}}{C + V_{s}}]$

V0 observer velocity

C sound velocity

Vs source velocity

$= \frac{\frac{6}{5} f_{0} - f_{0}}{f_{0}} * 100$

$= \frac{1}{5} * 100$

% change in frequency = 20%

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In a Young's double slit experiment, the separation between the slits is $0.15 m m$ . In the experiment, a source of light of wavelength $589 n m$ is used and the interference pattern is observed on a screen kept $1.5 m$ away. The separation between the successive bright fringes on the screen is:

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Contributor-Level 10

$\frac{Energy}{Volume} = \frac{{M L}^{2} T^{- 2}}{L^{3}} = ({M L}^{- 1} T^{- 2})$

The distance between two successive bright fringes is fringe width $(β)$ .

$β = \frac{λ D}{d} = \frac{589 * 10^{- 9} * 1.5}{0.15 * 10^{- 3}} = 5.9 m m$

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An unpolarised light beam of intensity 2l₀ is passed through a polaroid P and then through another polaroid Q which is oriented is such a way that its passing axis makes an angle of 30° relative to that of P. The intensity of the emergent light is

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Contributor-Level 10

$I = I_{0} c o s^{2} 30 °$

$= I_{0} {(\frac{\sqrt[]{3}}{2})}^{2} = \frac{3}{4} I_{0}$

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Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a * 10^‑2 W/m². The value of 'a' will be________.

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Contributor-Level 10

P = 10% of 110W = 11W

(Visible)

$I_{1} - I_{2} = \frac{ρ}{4 π} (\frac{1}{r_{1}^{2}} - \frac{1}{r_{2}^{2}})$

$= \frac{11}{4 * \frac{22}{7}} (\frac{1}{1} - \frac{1}{25})$

$\frac{7}{8} * (\frac{24}{25}) = a * 1 0^{- 2} W / m^{2}$

$\frac{28 * 24}{8} * 1 0^{- 2} = a * 1 0^{- 2}$

=84 * 10^-2 = a * 10^-2 = a = 84

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Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are p/2 and p/3 at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be________.

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Payal Gupta

Contributor-Level 10

$?_{A} = \frac{π}{2}, ?_{B} = \frac{π}{3}$

$I_{A} = I_{1} + I_{2} + 2 \sqrt[]{l_{1} l_{2}} c o s ? = l + 4 l + 2 \sqrt[]{l * 4 l} c o s \frac{π}{2} = 5 l$

$I_{B} = I + 4 I + 2 \sqrt[]{l . 4 l} c o s \frac{π}{3} = 5 l + 2 l = 7 l$

$Δ l = 7 l - 5 l = 2 l \Rightarrow x = 2$

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3 months ago

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Payal Gupta

Contributor-Level 10

P = 10% of 110W = 11W

(Visible)

$I_{1} - I_{2} = \frac{ρ}{4 π} (\frac{1}{r_{1}^{2}} - \frac{1}{r_{2}^{2}})$

$= \frac{11}{4 * \frac{22}{7}} (\frac{1}{1} - \frac{1}{25})$

$\frac{7}{8} * (\frac{24}{25}) = a * 1 0^{- 2} W / m^{2}$

$\frac{28 * 24}{8} * 1 0^{- 2} = a * 1 0^{- 2}$

⇒ 84 * 10^-2 = a * 10^-2 ⇒ a = 84

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3 months ago

Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is π. The difference of resultant intensities, at the point A and B, will be_________I.

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Contributor-Level 10

$I_{A} = 9 l + 4 l + 2 \sqrt[]{9 l * 4 l} c o s 0 °$

$= 13 l + 12 l = 25 l$

$I_{B} = 9 l + 4 l + 2 \sqrt[]{9 l * 4 l} c o s π$

$= 13 l - 12 l = l$

$∴ I_{A} - I_{B} = 24 l$

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In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is__________nm.

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