Question

If an object is placed at `A(OAgtf),` where f is the focal length of the lens, the image is formed at B. A perpendicular is erected at O and C is chosen on it such that the angle `(angleBCA)` is a right angle. Then, the value of f will be
A. `(AB)/(OC^2)`
B. `((AC)(BC))/(OC)`
C. `(OC^2)/(AB)`
D. `((OC)(AB))/(AC+BC)`

Answer 1

Correct Answer - C
`1/f=1/(OB)-1/(-OA)=1/(OB)+1/(OA)`
`f=((OA)(OB))/(AB+OB)`
`:. f=((OA)(OB))/(AB)….(i)`
Now, `AB^2=AC^2+BC^2`
or `(OA+OB)^2=AC^2+BC^2`
or `OA^2+OB^2+2(OA)(OB)=AC^2+BC^2`
`:. (AC^2-OC^2)+(BC^2-OC^2)`
`+2(OA)(OB)=AC^2+BC^2`
Solving, we get
`(OA)(OB)=OC^2`
Substituting in Eq. (i), we get
`f=(OC^2)/(AB)`