$A$ convex lens of focal length $12 \ cm$ is made of glass of refractive index $\mu_g = \frac{3}{2}$. What will be its focal length when immersed in a liquid of refractive index $\mu_l = \frac{5}{4}$? (in $cm$)

$A$ thin biconvex lens is made of glass $(\mu = 1.50)$ and both surfaces have a radius of curvature of $20 \ cm$. $A$ beam of incident light is parallel to the principal axis of the lens. The lens converges it at a distance $L \ cm$. Then $L = \dots$

$A$ diminished image of an object is to be obtained on a large screen $1\, m$ from it. This can be achieved by

$A$ thin convex lens made from crown glass $\left( \mu = \frac{3}{2} \right)$ has focal length $f$. When it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3}$,it has the focal lengths $f_1$ and $f_2$ respectively. The correct relation between the focal lengths is:

$A$ concave lens of glass,refractive index $1.5$,has both surfaces of same radius of curvature $R$. On immersion in a medium of refractive index $1.75$,it will behave as a

$A$ convex lens focuses an object $20 \ cm$ from it on a screen placed $5 \ cm$ away from it. $A$ glass plate (refractive index $\mu = \frac{7}{5}$) of thickness $t = 1.4 \ cm$ is inserted between the lens and the screen. What is the new distance of the object from the lens so that its image is again focused on the screen (in $cm$)?