The radii of curvature of both the surfaces of a convex lens are equal. One of the surfaces is made plane by grinding. What is the new focal power of the lens ?

Options

A. `P'=2`
B. `P'=(P)/(2)`
C. `P'=sqrt((P)/(2))`
D. `P'=(2)/(3)P`

Answer

B

Explanation

For the convex lens, `(1)/(f)=(n-1)((1)/(R_(1))-(1)/(R_(2)))`
But `R_(1)=R_(2)=R and R_(2)` is -ve
`therefore (1)/(f)=(n-1)((1)/(R )+(1)/(R )) therefore (1)/(f)=(2(n-1))/(R )`
`therefore f=(R )/(2(n-1)) " " ` …(1)
When it is converted into a plano convex lens,
`R_(1)=R, R_(2)=oo and (1)/(R_(2))=0`
and `(1)/(f')=(n-1)((1)/(R ))=(n-1)/(R )`
`therefore f'=(R )/((n-1))=2[(R )/(2(n-1))]=2f`
Thus the focal length is doubled and since `P=(100)/(f)`, the focal power is halved. Thus, `P'=(P)/(2)`.