A convex lens of refractive index 1.5 and focal length f=18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is f. The value of is \_ \_ \_ \_…

Question

A convex lens of refractive index 1.5 and focal length f=18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is f. The value of is \_ \_ \_ \_ . (refractive index of water =4/3 )

Accepted Answer

Correct answer: 3. (3) Using the lensmaker's equation, in air: 1f_ air= (n-1 )( 1R_1- 1R_2) where (n-1 )=0.5 for glass with n=1.5. In water: 1f_ water=( nn_ water-1)( 1R_1- 1R_2) =( 1.54/3-1)( 1R_1- 1R_2) = 18( 1R_1- 1R_2)Since ( 1R_1- 1R_2)= 2f_ air= 218, We get 1f_ water= 18 218= 172, so f_ water=72 cm. The difference is f_ water-f_ air=72-18=54 cm. Therefore = 5418=3.

$A convex lens of refractive index 1.5 and focal length is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is . The value of is . (refractive index of water )$

Solution

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