(1)/(F)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))^(3+sqrt(48)Gamma^(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 The focla lengh of a lens does not depend on the medium in which it is submerged. STATEMENT 2 (1)/(f)=(mu_(2)-mu_(1))/(mu_(1))((1)/(R_(1))-(1)/(R_(2)))

STATEMENT- 1 The focal length of a lens does not depend on the medium in which it is submerged. STATEMENT- 2 (1)/(f)=(mu_(2)-mu_(1))/(mu_(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)))

(1)/(r^(2))+(1)/(r_(1)^(2))+(1)/(r_(2)^(2))+(1)/(r_(3)^(2)) equals

((1)/(r)+(1)/(r_(1))+(1)/(r_(2))+(1)/(r_(3)))^(2)=(4)/(r)((1)/(r_(1))+(1)/(r_(2))+(1)/(r_(3)))

Prove that ((1)/(r)-(1)/(r_(1)))((1)/(r)-(1)/(r_(2)))((1)/(r)-(1)/(r_(3)))=(abc)/(Delta^(3))=(4R)/(r^(2)s^(2))

Show that ((1)/(r _(1))+ (1)/(r_(2)))((1)/(r_(2))+(1)/(r _(3)))((1)/(r _(3))+(1)/(r_(1)))=(64 R^(3))/(a ^(2)b^(2) c^(2))