In the given figure, S is a monochromatic point source emitting light of wavelength $λ = 500 nm$ . A thin lens of circular shape and focal length 0.1m is cut into two identical halves L₁ and L₂ 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 L₁and L₂ is 0.15m, while that from L₁ and L₂ 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?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

1 mm

2 mm

2.5 mm

1.5 mm

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The images of S due to both parts of lenses are
formed at S ₁and S₂ whose distance ‘v’ from the lens is give n by
$\begin{array}{l} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{0 .1} = \frac{1}{v} - \frac{1}{(- 0 .15)} \\ \frac{1}{v} = \frac{1}{0 .1} - \frac{1}{0 .15} = \frac{10}{1} - \frac{100}{15} = \frac{150 - 100}{15} = \frac{50}{15} = \frac{10}{3} \\ v = \frac{3}{10} = 0 .3 m \end{array}$
From similar $Δ_{les} {SS}_{1} S_{2} and {SP}_{1} P_{2}$
$\begin{array}{l} \frac{d}{0 .5 \times 10^{- 3}} = \frac{0 .3 + 0 .15}{0 .15} \\ d = \frac{0 .45}{0 .15} \times 0 .5 \times 10^{- 3} = 1 .5 \times 10^{- 3} m \end{array}$
For third maximum
$\begin{array}{l} y_{n} = \frac{3 λD}{d} = \frac{3 \times 500 \times 10^{- 9} \times 1}{1 .5 \times 10^{- 3}} = 10^{- 9} \times 10^{6} \\ y_{n} = 10^{- 3} m \Rightarrow y_{3} = 1 mm \end{array}$