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Q.

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

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$f = \frac{R}{μ - 1} = \frac{30}{1.75 - 1} = \frac{30}{(3 / 4)} = 40 cm$
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

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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.