A meniscus lens is made of a material of refractive index -2- Both its surfaces have radii of curvature R- It has two different media of refractive indices -1 and -3 respectively- on its two sides see figure- Calculate its focal length for -1 - -2 - -3- when light is incident on it as shown-

Apply the lens makers formula P_lens = ( n_lens n_0)n_0( 1R_1 1R_2) All details are provided in the question. R_1 = 15cm;R_2 = #160; 15cm;n( len ...

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Byju's Answer

Standard XII

Physics

Power and Thin Lenses in Contact

A meniscus le...

Question

A meniscus lens is made of a material of refractive index $μ_{2}$ . Both its surfaces have radii of curvature $R$ . It has two different media of refractive indices $μ_{1}$ and $μ_{3}$ respectively, on its two sides (see figure). Calculate its focal length for $μ_{1} < μ_{2} < μ_{3}$ , when light is incident on it as shown.

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Solution

Apply the lens makers formula
$P_{l e n s} = \frac{(n_{l e n s} - n_{0})}{n_{0}} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
All details are provided in the question.
$R_{1} = 15 c m; R_{2} = - 15 c m; n (l e n s) = 1.6; n (0) = 1.65$
$F o c a l l e n g t h = \frac{1}{P (l e n s)} = - 247 c m (a p p r o x)]$

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