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Physics Question on Youngs double slit experiment

White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and the screen is at a distance d (>> b) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelengths are

A

λ=b2/d\lambda=b^2/d

B

λ=2b2/d\lambda=2b^2/d

C

λ=b2/3d\lambda=b^2/3d

D

λ=2b2/3d\lambda=2b^2/3d

Answer

λ=b2/3d\lambda=b^2/3d

Explanation

Solution

At P (directly infront of S1S_1 ) y = b / 2
\therefore Path difference,
ΔX=S2PS1P=y.(b)d=(b2)(b)d=b22d\Delta X=S_2 P-S_1 P=\frac{y.(b)}{d}=\frac{\big(\frac{b}{2}\big)(b)}{d}=\frac{b^2}{2d}
Those wavelengths will be missing for which
\hspace25mm \Delta X=\frac{\lambda_1}{2},\frac{3 \lambda_2}{2},\frac{5 \lambda_3}{2}...
\therefore \hspace25mm \lambda_1=2 \Delta x=\frac{b^2}{d}
\hspace25mm \lambda_2=\frac{2 \Delta x}{3}=\frac{b^2}{3d}
\hspace25mm \lambda_3=\frac{2 \Delta x}{5}=\frac{b^2}{5d}
\therefore Correct options are (a) and (c).