Question

The radius of curvature of each surface of an equiconvex lens is R = 42 cm. Refractive index of the glass = 1.25. If the final image forms after four internal reflections (but not TIR) inside the lens (for paraxial ray incident beam parallel to principal axis) calculate the distance of the image from the lens, in cm.

Answer 1

Given : The radius of curvature of each surface of an equiconvex lens is R = 42 cm. Refractive index of the glass = 1.25. If the final image forms after four internal reflections (but not TIR) inside the lens (for paraxial ray incident beam parallel to principal axis).

To find : The distance of the image from the lens.

solution : Let μ is refractive index of lens.

n is number we of internal reflections.

R is radius of curvature of lens.

formula of distance of the image from the lens is given by, $v=\frac{R}{2(n\mu+\mu-1)}$

here R = 42 cm, n = 4 , μ = 1.25

so, v = 42/2{(4 × 1.25 + 1.25 - 1)}

= 42/2(4 + 1.25)

= 42/10.5

= 4 cm

therefore image distance from the lens is 4 cm

Answer 2

Given:

Radius,

R = 42 cm

Number of internal reflections,

n = 4

Refractive index of lens,

μ = 1.25

To Find:

Distance of image = ?

Solution:

As we know,

The formula of distance will be:

⇒  $v=\frac{R}{2(n \mu+\mu-1)}$

On substituting the values, we get

⇒     $=\frac{42}{2(4\times 1.25 + 1.25 - 1)}$

⇒     $=\frac{42}{2(4 + 1.25)}$

⇒     $=\frac{42}{10.5}$

⇒     $=4 \ cm$

So that the image distance will be "4 cm".