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Correct option is D)

Here the light passes through three refracting media, $I, II & III$ as shown in the figure.

Media $I & II$ has a refractive index $1.5$ and radius of curvature $20cm$. While medium $II$ has a refractive index of $1.7$ and it acts as a concave lens.

Let the focal length of the three media be $f_{1},f_{2},f_{3}$.

Applying the lens makers formula

$f1 =(μ−1)(R_{1}1 −R_{2}1 )$

$R_{1}=∞$ since it is a plano convex lens.

$f_{1}1 =(1.5−1)(∞1 −−201 )=0.5×201 =401 $

$f_{2}1 =(1.7−1)(−201 −201 )=0.7×(−202 )=−100.7 $

Since media $I & III$ are similar we can write

$f_{3}1 =401 $

$f_{eq}1 =f_{1}1 +f_{2}1 +f_{3}1 $

$⇒f_{eq}1 =401 −100.7 +401 $

$⇒f_{eq}1 =401−2.8+1 =−400.8 =−501 $

$∴f_{eq}=−50cm$

Hence the correct answer is option (D).

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