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JEE Main 202429 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2024 · 29 January · Evening Shift · Physics · Wave Optics

In a single slit diffraction pattern, a light of wavelength 6000 o...

In a single slit diffraction pattern, a light of wavelength 6000Ao6000\overset{o}{A} is used. The distance between the first and third minima in the diffraction pattern is found to be 3mm3mm when the screen is placed 50cm50cm away from slits. The width of the slit is ____×10−4m\times {10}^{- 4}m.In a single slit diffraction pattern, a light of wavelength 6000 o A is used. The distance between the first and third

Correct answer

0

Step-by-step solution

For nth{n}^{th} minima:

⇒bsinθ=nλ⇒ bsin\theta = n\lambda

(λ\lambda is very small, so sinθsin\theta is very small, hence sinθ≃tanθsin\theta ≃ tan\theta.

⇒btanθ=nλ⇒ btan\theta = n\lambda

⇒byD=nλ⇒ b\dfrac{y}{D} = n\lambda

⇒yn=n\lambdaDb⇒ {y}_{n} = \dfrac{n\lambdaD}{b} (Position of nth{n}^{th} minima)

In a single slit diffraction pattern, a light of wavelength 6000 o A is used. The distance between the first and third

B→1stB \to {1}^{\text{st}} minima and A→3rdA \to {3}^{\text{rd}}:

y3=3\lambdaDb,y1=\lambdaDb{y}_{3} = \dfrac{3\lambdaD}{b}, {y}_{1} = \dfrac{\lambdaD}{b}

⇒\Deltay=y3−y1=2\lambdaDb⇒ \Deltay = {y}_{3} - {y}_{1} = \dfrac{2\lambdaD}{b}

⇒3×10−3=2×6000×10−10×0.5b⇒ 3 \times {10}^{- 3} = \dfrac{2 \times 6000 \times {10}^{- 10} \times 0 . 5}{b}

⇒b=2×6000×10−10×0.53×10−3⇒ b = \dfrac{2 \times 6000 \times {10}^{- 10} \times 0 . 5}{3 \times {10}^{- 3}}

⇒b=2×10−4m⇒ b = 2 \times {10}^{- 4}m

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