SATHEE: Optics Question 158

Optics Question 158

Question: Two ideal slits S1 and S2 are at a distance d apart, and illuminated by light of wavelength l passing through an ideal source slit S placed on the line through S2 as shown. The distance between the planes of slits and the source slit is D. A screen is held at a distance D from the plane of the slits. The minimum value of d for which there is darkness at O is

Options:

A) $\sqrt{\frac{3 λ D}{2}}$

B) $\sqrt{λ D}$

C) $\sqrt{\frac{λ D}{2}}$

D) $\sqrt{3 λ D}$

Show Answer

Answer:

Correct Answer: C

Solution:

Path difference between the waves reaching at $P,$

$Δ = Δ_{1} + Δ_{2}$ where $Δ_{1} =$

Initial path difference $Δ_{2} =$

Path difference between the waves after emerging from slits. $Δ_{1} = S S_{1} - S S_{2} = \sqrt{D^{2} + d^{2}} - D$

and $Δ_{2} = S_{1} O - S_{2} O = \sqrt{D^{2} + d^{2}} - D$

$∴ Δ = 2 {(D^{2} + d^{2})}^{\frac{1}{2}} - D = 2 (D^{2} + \frac{d^{2}}{2 D}) - D$

$= \frac{d^{2}}{D}$

(From Binomial expansion) For obtaining dark at O, $Δ$ must be equals to $(2 n - 1) \frac{λ}{2}$ i.e. $\frac{d^{2}}{D} = (2 n - 1) \frac{λ}{2} \Rightarrow d \sqrt{\frac{(2 n - 1) λ D}{2}}$

For minimum distance $n = 1$ so $d = \sqrt{\frac{λ D}{2}}$

Optics Question 158

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Optics Question 158

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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