A convex lens (n = 1.5) of focal length `f_a` is immersed in water n = 1.33.

A convex lens (n = 1.5) of focal length f_a is immersed in carbon disulphide n = 1.6, how does the lens behave in the two cases?

A convex lens of focal length 24 cm is totally immersed in water (n = 1.33). Find its focal length in water (n = 1.5).

A thin lens of focal length + 12 cm is immersed in water (mu = 1.33). What is its new focal length ?

A thin lens of focal length + 12 cm is immersed in water (mu = 1.33). What is its new focal length ?

A thin lens of focal length + 12 cm is immersed in water (mu = 1.33). What is its new focal length ?

A convex mirror of focal length f (in air) is immersed in water (mu = 4/3) . The focal length of the mirror in water will be -

Assertion : A double convex lens (mu=1.5) has focal length 10 cm . When the lens is immersed in water (mu=4//3) its focal length becomes 40 cm . Reason : (1)/(f)=(mu_(1)-mu_(m))/(mu_(m))((1)/(R_(1))-(1)/(R_(2)))

Assertion : A double convex lens (mu=1.5) has focal length 10 cm . When the lens is immersed in water (mu=4//3) its focal length becomes 40 cm . Reason : (1)/(f)=(mu_(1)-mu_(m))/(mu_(m))((1)/(R_(1))-(1)/(R_(2)))