A concave lens has same radii of curvature for both sides and is made of material of refractive in index `1.6` It is immersed in a liquid of `mu = 1.4` Calculate ratio of focal lengths of lens in air and liquid.

A corverging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6 . If is immersed in a liquid of refractive index 1.3 , what will be its new focal length ?

A converging lens has focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, what will be its new focal length?

A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, what will be the new focal length? How does the nature of the lens change. If this lens is immersed is a liquid of refractive index 1.8 ?

A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6 . If it is immersed in a liquid of refractive index 1.3 , what will be its new foacl length ?

A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6 . If it is immersed in a liquid of refractive index 1.3 , what will be its new foacl length ?

A biconvex lens of focal length f and radii of curvature of both the surface R is made of a material of refractive index n_(1) . This lens is placed in a liquid of refractive index n_(2) . Now this lens will behave like

A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in water of refractive index 1.3, what will be the new focal length?