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(D) For the image to be formed at infinity,the object must be placed at the principal focus of the convex lens.The original distance of the object fro

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$15$

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(D) For the image to be formed at infinity,the object must be placed at the principal focus of the convex lens.The original distance of the object from the lens is $u = 25 \, cm$. The focal length of the lens is $f = 20 \, cm$.When a glass slab of thickness $t$ and refractive index $\mu = 1.5$ is introduced,the object appears to be shifted towards the lens by a distance $x$ given by:$x = t \left( 1 - \frac{1}{\mu} \right)$Substituting the values:$x = t \left( 1 - \frac{1}{1.5} \right) = t \left( 1 - \frac{2}{3} \right) = \frac{t}{3}$The new effective distance of the object from the lens becomes $u' = u - x = 25 - \frac{t}{3}$.Since the image is formed at infinity,the effective object distance must be equal to the focal length:$25 - \frac{t}{3} = 20$Solving for $t$:$\frac{t}{3} = 25 - 20 = 5$$t = 15 \, cm$.

$A$ point object is placed at a distance of $25 \, cm$ from a convex lens of focal length $20 \, cm$. If a glass slab of thickness $t$ and refractive index $1.5$ is inserted between the lens and the object,the image is formed at infinity. The thickness $t$ is....$cm$.

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