Question

a point source is located at a distance of 20cm from the front surfaceof a gllass bi convex lens.the lens has thickness of 5 cm. the refractive index of glass is 1.5. the distance of image formed by it from the rear surface of this lens is

Answer 1

We need the radii of surfaces of lens R1 and R2.  I am giving the formulas and the procedure here.  So follow these steps.

1) Formulas:

Back Effective focal length to focus F' = B EFL = f '  (+ve)
Effect focal length in front of the front surface =  EFL = f (-ve)

       R1 = +ve          R2 = -ve
Thickness of the lens = d
μ = refractive index of the lens wrt air (medium) around the lens on both sides

Frontal focal length (distance) from surface to Focus F = FFL = -ve
Back Focal length (distance) from back surface to Focus F' = BFL = + ve

$\frac{1}{EFL}=\frac{1}{f}=(\mu - 1) [\frac{1}{R_1}-\frac{1}{R_2}]+\frac{(\mu -1)^2d}{\mu R_1 R_2}\\\\\frac{1}{BEFL}=\frac{1}{f'}=(\mu - 1) [\frac{1}{R_2}-\frac{1}{R_1}]+\frac{(\mu -1)^2d}{\mu R_1 R_2}\\\\FFL= f [1+\frac{(\mu-1)d}{\mu\ R_2}].\\\\BFL= f [1-\frac{(\mu-1)d}{\mu\ R_1}].$

2. Given values:

Given R1 = R2,  d = 5 cm,  μ = 1.5 ,  u = - 20 cm 

3. Procedure:

 1)  Find  f  =  f'
 2)  Then  find v from the formula 1/f = 1/v - 1/u
 3)  From   v now subtract  (f'  -  BFL)  to get the distance of image from the rear surface.  This value of (f ' -  BFL)  or  (f  -  FFL)  may be close to  d/2.