" 13.Establish the formula "(1)/(f)=(mu-1)((1)/(r_(1))-(1)/(r_(2)))

Statement-1 : Focal length of an equiconvex lens of mu = 3//2 is equal to radius of curvature of each surface. Statement-2 : it follows from (1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))

Statement-1 : A glass convex lens (refractive index, mu_(1)=1.5 ), when immersed in a liquid of refractive index mu_(2)=2 , functions as a diverging lens. Statement-2 : The focal length of a lens is given by (1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2))) , where mu is the relative refractive index of the material of the lens with respect to the surrounding medium.

Statement-1 : A glass convex lens (refractive index, mu_(1)=1.5 ), when immersed in a liquid of refractive index mu_(2)=2 , functions as a diverging lens. Statement-2 : The focal length of a lens is given by (1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2))) , where mu is the relative refractive index of the material of the lens with respect to the surrounding medium.

STATEMENT- 1 Power of a lens depends on nature of material of lens, medium in which it is placed and radii of curvature of its surface. STATEMENT 2 it follows the relation p=(1)/(f)=((mu_(2))/(mu_(1))-1)[(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)))

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)))

Derive expression for the lens maker’s formula i.e.: frac(1)(f)= (mu-1) (frac(1)(R_1)-frac(1)(R_2)) where the letters have their usual meanings