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In the given figure, S is a monochromatic point source emitting light of wavelength $λ = 500 n m .$ A thin lens of circular shape and focal length 0.1m is cut into two identical halves $L_{1} a n d L_{2}$ by a plane passing through diameter. The two halves are symmetrically placed about the central axis SO with gap of 0.5mm. The distance along the axis the from S to $L_{1} a n d L_{2}$ is 0.15m, while that from $L_{1} a n d L_{2}$ to ‘O’is 1.3m. The screen at O is normal to SO. If the third maxima occurs at point A on the screen find the distance OA?(in mm)

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Detailed Solution

$∵ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} ϑ = 0.3 m \frac{0.5 \times 10^{- 3}}{0.15} = \frac{d}{0.45} d = 1.5 \times 10^{- 3} m y = \frac{3 λ D}{d} = \frac{3 \times 500 \times 10^{- 10} \times 1}{1.5 \times 10^{- 3}} y = 10^{- 3} m = 1 m m$

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