SATHEE: Optics Question 901

Optics Question 901

Question: A parallel beam of light $(λ = 5000 \overset{o}{A})$ is incident at an angle $α = 30^{\circ}$ with the normal to the slit plane in YDSE. Assume that the intensity due to each slit at any point on the screen is $I_{0}$ . Point O is equidistant from $S_{1}$ and $S_{2}$ . The distance between slit is 1 mm, then the intensity at

Options:

A) O is $3 I_{0}$

B) O is zero

C) a point 1 m below O is $4 I_{0}$

D) a point on the screen 1 m below O is zero

Show Answer

Answer:

Correct Answer: C

Solution:

[c] The path difference at O, $Δ x = d \sin α = d \sin 30^{o} = \frac{10^{- 3}}{2} m$

Now $ϕ = \frac{2 π}{λ} . Δ x = \frac{2 π}{5000 \times 10^{- 10}} \times \frac{10^{- 3}}{2} = 2 π \times 10^{3}$

So $I = I_{o} + I_{o} + 2 \sqrt{I_{o} I_{o}} \cos (2 π \times 10^{3}) = 4 I_{o}$

The angular position of P, $\tan θ = \frac{1}{\sqrt{3}}$ ; or $θ = 30^{o}$ .

It means path $d i f f^{n}$ at P is also $Δ x$

and hence $I = 4 I_{o}$

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Optics Question 901

Question: A parallel beam of light (λ=5000Ao) is incident at an angle α=30∘ with the normal to the slit plane in YDSE. Assume that the intensity due to each slit at any point on the screen is I0 . Point O is equidistant from S1 and S2 . The distance between slit is 1 mm, then the intensity at

Options:

Answer:

Solution:

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