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)
Answer
3
Solution
Lens maker's formula: 1f = ( l m - 1 ) ( 1R1 - 1R2 )
In air: 1f = (1.5 - 1)K = 0.5K
In water: 1fw = ( 1.54/3 - 1 )K = 18K
fwf = 0.51/8 = 4, fw = 4f = 72 cm
fw - f = 4f - f = 3f
= 3
Answer: 3