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Two coherent sources have intensity in the ratio of $\frac{100}{1}$ . Ratio of (intensity) max/(intensity) min is

$\frac{1}{100}$

$\frac{1}{10}$

$\frac{10}{1}$

$\frac{3}{2}$

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Explanation

$\frac{I_{1}}{I_{2}} = \frac{100}{1}$

Now $\frac{I_{\max}}{I_{\min}} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\sqrt{\frac{I_{1}}{I_{2}}} - 1})}^{2} = {(\frac{\sqrt{100} + 1}{\sqrt{100} - 1})}^{2} = \frac{121}{81} \approx \frac{3}{2}$

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If two waves represented by $y_{1} = 4 \sin ω t$ and $y_{2} = 3 \sin (ω t + \frac{π}{3})$ interfere at a point, the amplitude of the resulting wave will be about

3.5

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Explanation

$ϕ = π / 3, a_{1} = 4, a_{2} = 3$

So, $A = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} . a_{2} \cos φ} \Rightarrow A \approx 6$

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Two coherent sources of intensities, I₁ and I₂ produce an interference pattern. The maximum intensity in the interference pattern will be

I₁ + I₂

$I_{1}^{2} + I_{2}^{2}$

(I₁ + I₂)²

${(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$

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Explanation

Resultant intensity $I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$

For maximum $I_{R}, ϕ = 0^{o}$

⇒ $I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$

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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 beams is $\frac{π}{2}$ at point A and π at point B. Then the difference between the resultant intensities at A and B is

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Explanation

Resultant intensity, at point A,

$I_{A} = I_{1} + I_{2} = 5 I;$

And, at point B,

$I_{B} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos π = 5 I - 4 I$

$I_{B} = I$

So, $I_{A} - I_{B} = 4 I$

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Two waves are represented by the equations $y_{1} = a \sin ω t$ and $y_{2} = a \cos ω t .$ The first wave

Leads the second by $π$

Lags the second by $2 π$

Leads the second by $\frac{π}{2}$

Lags the second by $\frac{π}{2}$

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Explanation

$y_{1} = a \sin ω t, y_{2} = a \cos ω t = a \sin (ω t + \frac{π}{2})$

Hence the first wave lags the second by $\frac{π}{2}$

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If an interference pattern have maximum and minimum intensities in 36 : 1 ratio then what will be the ratio of amplitudes

5 : 7

7 : 4

4 : 7

7 : 5

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Explanation

$\frac{I_{\max}}{I_{\min}} = {(\frac{\frac{a_{1}}{a_{2}} + 1}{\frac{a_{1}}{a_{2}} - 1})}^{2} \Rightarrow \frac{a_{1} + a_{2}}{a_{1} - a_{2}} = 6$

$\frac{a_{2}}{a_{1}} = 7 : 5$

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In a certain double slit experimental arrangement interference fringes of width 1.0 mm each are observed when light of wavelength 5000 Å is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength 6000 Å, the fringe width will be

0.5 mm

1.0 mm

1.2 mm

1.5 mm

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Explanation

$\frac{β_{1}}{β_{2}} = \frac{λ_{1}}{λ_{2}}$ or $\frac{1 .0}{β_{2}} = \frac{5000}{6000}$ or $β_{2} = \frac{6000}{5000} = 1 .2 m m$ .

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Two coherent light sources S₁ and S₂ (λ= 6000 Å) are 1mm apart from each other. The screen is placed at a distance of 25 cm from the sources. The width of the fringes on the screen should be

0.015 cm

0.025 cm

0.010 cm

0.030 cm

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Explanation

$β = \frac{6000 \times 10^{- 10} \times 25 \times 10^{- 2}}{10^{- 3}}$

$= 150000 \times 10^{- 9} = 0 .15 \times 10^{- 3} m = 0 .015 c m$ .

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The Young's experiment is performed with the lights of blue (λ = 4360 Å) and green colour (λ = 5460 Å), If the distance of the 4th fringe from the centre is x, then

x (Blue) = x (Green)

x (Blue)> x (Green)

x (Blue) < x (Green)

$\frac{x (B l u e)}{x (G r e e n)} = \frac{5460}{4360}$

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Explanation

Distance of n^th bright fringe $y_{n} = \frac{n λ D}{d} i . e . y_{n} \propto λ$

$∴ \frac{x_{n_{1}}}{x_{n_{2}}} = \frac{λ_{1}}{λ_{2}} \Rightarrow \frac{x (B l u e)}{x (G r e e n)} = \frac{4360}{5460}$

∴ x (Green) > x (Blue).

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In Young's double slit experiment, if L is the distance between the slits and the screen upon which interference pattern is observed, x is the average distance between the adjacent fringes and d being the slit separation. The wavelength of light is given by

$\frac{x d}{L}$

$\frac{x L}{d}$

$\frac{L d}{x}$

$\frac{1}{L d x}$

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Explanation

We know that fringe width $β = \frac{D λ}{d}$

$∴ x = \frac{L λ}{d} \Rightarrow λ = \frac{x d}{L}$

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