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 of focal length 0.1m is cut into two identical halves L₁ and L₂ by a plane passing through to a diameter The two halves are symmetrically placed about the central axis SO with a gap of 0.5mm as shown in fig. The screen is at O normal to SO. If the third intensity maximum occurs at point A on the screen find the distance OA in cm.

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Detailed Solution

The images of S due to both parts of lenses are formed at $S_{1} a n d S_{2}$ whose distance ‘v’ from the lens is give n by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{0.1} = \frac{1}{v} - \frac{1}{(- 0.15)}$

$\frac{1}{v} = \frac{1}{0.1} - \frac{1}{0.15} = \frac{10}{1} - \frac{100}{15} = \frac{150 - 100}{15} = \frac{50}{15} = \frac{10}{3} v = \frac{3}{10} = 0.3 m$

From similar $Δ_{l e s} S S_{1} S_{2} a n d S P_{1} P_{2}$

$\frac{d}{0.5 \times 10^{- 3}} = \frac{0.3 + 0.15}{0.15}$

$d = \frac{0.45}{0.15} \times 0.5 \times 10^{- 3} = 1.5 \times 10^{- 3} m y_{n} = \frac{3 λ D}{d} = \frac{3 \times 500 \times 10^{- 9} \times 1}{1.5 \times 10^{- 3}} = 10^{- 9} \times 10^{6} y_{n} = 10^{- 3} m \Rightarrow y_{3} = 1 m m = 0.1 c m$