Question

The focal length of a this lens in vacuum is f. If the material of the lens has µ = 3/2, its focal length when immersed in water refractive index 4/3 will

विकल्प

• F

• 4f/3

• 3f

• 4f

Answer 1

4f

Explanation:

`1/(Fa) = ((ug)/(ua) - 1) (1/R_1 - 1/R_2)`

`1/(Fw) = ((ug)/(muw) - 1) (1/R_1 - 1/R_2)`

∴ Dividing (i) by (ii) we get

`(fw)/(fa) = (((mug)/(mua - 1)))/(((mug)/(muw - 1))`

= `((3/(2 - 1)))/(((3/2)/(4/2) = 1)`

= `(1/2)/(1/0)` = 4

fw = 4fa = 4f