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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
you need to give the radii of curvature of the surfaces of the lens



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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.

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