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Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsYDSEEasy2 minQB

PhysicsEasysingle choice

In Young’s double-slit experiment, the slit separation is $1 mm$ and the screen is at a distance of $1 m$ from the slit. For a monochromatic light of wavelength $500 nm$ , the distance of $3^{rd}$ minima from the central maxima is

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Answer:

Solution:

Distance of n^thminima from central bright fringe

$x_{n} = \frac{(2 n - 1) λ D}{2 d}$

For $n = 3$ i.e., 3^rdminima

$x_{3} = \frac{(2 \times 3 - 1) \times 500 \times 1 0^{- 9} \times 1}{2 \times 1 \times 1 0^{- 3}}$

$= \frac{5 \times 500 \times 1 0^{- 6}}{2} = 1.25 \times 1 0^{- 3} m = 1.25 mm$

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:YDSE

⏱ 2mℹ️ Source: QB

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