कथन 1. समोत्तल लेंस की फोकस - दूरी किसी वक्रता - त्रिज्या के बराबर होती हैं ।
कथन 2. `mu = 1.5 ` हो तथा `(1)/(f) = (mu -1) ((1)/(R_(1)) - (1)/(R_(2)))`.

Value of 1/(r_(1)^2)+ 1/(r_(2)^2)+ 1/(r_(3)^2)+ 1/(r_()^2) is :

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)))

The value of 1/(r_(1)^(2))+1/(r_(2)^(2))+1/(r_(2)^(3))+1/(r^(2)) , is

In a triangle ABC,if (1)/(r_(1)^(2))+(1)/(r_(2)^(2))+(1)/(r_(3)^(2))+(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)))

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.