In the given figure, the radius of curvature of curved surface for both the plano-convex and plano-concave lens is 10 cm and refractive index for both is 1.5. The location of the final image after all the refractions through lenses is

Question

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Answer 1

The focal length of the plano - convex lens is:
$\frac{1}{f _{1}} = (1.5 - 1) (\frac{1}{+ 1 0} - \frac{1}{\infty}) = \frac{1}{2 0}$
Focal length of plano - concave lens is:
$\frac{1}{f _{2}} = (1.5 - 1) (\frac{1}{\infty} - \frac{1}{1 0}) = - \frac{1}{2 0}$
since parallel beams are incident on the lens, its image from plano - concave lens will be formed at $+ 20 c m$ from it (at the focus) and will act as an object for the plano - concave lens. since the two lens are at a distance of $10 c m$ from each other, therefore, for the next lens
$u = + 10 c m$
$v = \frac{u f}{u + f} = \frac{1 0 \times 2 0}{1 0 - 2 0}$
$\Rightarrow v = (\frac{2 0 0}{1 0}) c m = 20 c m$
Hence, the correct option is $(B)$

Answer 2

The focal length of the plano - convex lens is:

\frac{1}{f _{1}} = (1.5 - 1) (\frac{1}{+ 1 0} - \frac{1}{\infty}) = \frac{1}{2 0}

Focal length of plano - concave lens is:

\frac{1}{f _{2}} = (1.5 - 1) (\frac{1}{\infty} - \frac{1}{1 0}) = - \frac{1}{2 0}

since parallel beams are incident on the lens, its image from plano - concave lens will be formed at

+ 20 c m

from it (at the focus) and will act as an object for the plano - concave lens. since the two lens are at a distance of

10 c m

from each other, therefore, for the next lens

u = + 10 c m

v = \frac{u f}{u + f} = \frac{1 0 \times 2 0}{1 0 - 2 0}

\Rightarrow v = (\frac{2 0 0}{1 0}) c m = 20 c m

Hence, the correct option is

(B)