Question

A point source is located at a distance of 20 cm from the the front surface of a bi-convex lens. The lens has a thickness of 5 cm and radius of curvature of it's surface is 5cm. 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

for first surface of bi-convex lens

u = -20cm, μ1 = 1, μ2 = 1.5 , R = 5cm

μ2/v - μ1/u = (μ2 - μ1)/R

⇒1.5/v - 1/-20 = (1.5 - 1)/5

⇒1.5/v + 1/20 = 1/10

⇒1.5/v = 1/20

⇒v = 30 cm

for second surface of bi-convex lens

u = 30 - 5 = 25cm, μ1 = 1.5 , μ2 = 1 and R = -5 cm

μ2/v - μ1/u = (μ2 - μ1)/R

⇒1/v - 1.5/25 = (1 - 1.5)/-5

⇒1/v - 3/50 = 1/10

⇒1/v = 8/50

⇒v = 50/8 = 6.25 cm

hence, The distance of image formed by it from the rear surface of this lens is 6.25 cm

Answer 2

$\huge\bold\purple{Answer:-}$

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.