Question

A biconvex lens sits between an object and a concave spherical mirror. The lens has a focal length f and the mirror has a radius of curvature of R. The object is a distance 2f from the lens and the distance from the lens to the mirror is 2f + R. The object is illuminated and light scattered from the object shines through the lens to the mirror, reflects, travels back through the lens, and forms a final image.

(a) Draw an accurate ray diagram for this arrangement using a ruler.

(b) What is the distance from the object to its final image?

(c) What is the lateral magnification?

(d) Is the image real or is it virtual?

(e) Is the image upright or inverted?

Answer 1

Using a mirror formula, we have

1 / d₀ + 1 / d_i = 1 / f

1 / d_i = 1 / (R/2) - 1 / (2f + R)

1 / d_i = (2/R) - 1 / (2f + R)

1 / d_i = 2 (2f + R) - R / R (2f + R)

1 / d_i = (4f + R) / (2Rf + R²)

d_i = (2Rf + R²) / (4f + R)

(b) The distance from the object to its final image will be given as :

D = d₀ + d_i

where, d₀ = object distance from the lens = 2f

d_i = image distance from the mirror = (2Rf + R²) / (4f + R)

D = 2f + [(2Rf + R²) / (4f + R)]

D = 2f (4f + R) + (2Rf + R²) / (4f + R)

D = (8 f² + 2Rf +2Rf + R²) / (4f + R)

D = (8 f² + 4Rf + R²) / (4f + R)

(c) The ratio of the height of an object to the height of the real image is known as the "Lateral magnification".

Formula for Lateral magnification is given as -

m = h_i / h₀

If m is negative, then the image is inverted which relative to the object.

(d) The image will be real.

(e) The image will be inverted.