Question

A turnip sits before a thin converging lens, outside the focal point of the lens. The lens is filled with a transparent gel so that it is flexible; by squeezing its ends toward its center [as indicated in figure (a)], you can change the curvature of its front and rear sides. Suppose that the image must be formed on a card which is at a certain distance behind the lens [figure (b)], while you move the turnip away from the lens, then you should

• A. decrease the squeeze of the lens
• B. increase the squeeze of the lens
• C. keep the card and lens as it is
• D. move the card away from the lens

Answer 1

The correct option is A decrease the squeeze of the lens

Using formula for lens:-

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

using sign convention we see that :

$v(+),u(-),f(+)$

so replacing $u->(-u)$

$\frac{1}{v}-\frac{1}{(-u)}=\frac{1}{f}$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

since image is formed at fixed distance $\frac{1}{v}$ is constant ,so increasing u will result in decrements of $\frac{1}{u}$ hence,

$f$ should be increased by decreasing squeeze of lens .