As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is $30 cm$ and refraction index of the material for both the lenses is $1.75$ . Both the lenses are placed at distance of $40 cm$ from each other. Due to the combination, the image of the object is formed at distance $x =_______cm$ from plano concave lens.

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answer is 120.

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Detailed Solution

$f = \frac{R}{μ - 1} = \frac{30}{1.75 - 1} = \frac{30}{(3 / 4)} = 40 cm$
Question Image
I₁ = V.I. formed by concave lens = R.O to the convex lens
Convex lens: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \frac{1}{v} - \frac{1}{- 80} = \frac{1}{40} \Rightarrow \frac{1}{V} = \frac{1}{40} - \frac{1}{80} = \frac{2 - 1}{80} = \frac{1}{80}$
$∴$ v = +80cm, I₂ = R.I of I₁ formed by convex lens, d = 40 + 80 = 120cm