Question

At what point on the principal axis of the two mirrors should a point source of light be placed for the rays of converge at the same point after being reflected first from the convex mirror and then from the concave mirror

Answer 1

At (f + f√3) point on the principal axis of the two mirrors should a point source of light from the convex mirror and (3f - f√3) from the concave mirror.

Explanation:

For convex lens, the focal length is given by the formula:

1/f = 1/v + 1/u

Where,

f = Focal length

u = Object distance = -d

v = Image distance = v'

Now, the formula becomes,

1/f = 1/v' + -1/d

1/v' = 1/f + 1/d

1/v' = (d + f)/fd

∴ v' = fd/(d + f)

For concave lens,

f = -4f

u = -(Distance between Q and Pole of concave mirror)

v = v'

Now,

u = -(v' + 4f)

On substituting 'v'', we get,

u = -((fd/(d + f)) + 4f)

∴ u = -(4f² + 5fd)/(f + d)

Now,

v = - (distance between O and pole of concave mirror)

v = -(4f - d)

Now, the focal length of concave lens is given as:

-1/f = (-(4f² + 5fd)/(f + d)) - (1/(4f - d))

1/f = ((4f² + 5fd)/(f + d)) + (1/(4f - d))

((4f² + 5fd)/(f + d)) = 1/f - (1/(4f - d))

((4f² + 5fd)/(f + d)) = (4f² - d)/(3f - d)

(f + d)/(4f² + 5fd) = (3f - d)/(4f² - d)

(f + d) × (4f² - d) = (3f - d) × (4f² + 5fd)

4f³ + 4f²d - f²d - fd² = 12f³ - 4f²d + 15f²d - 5fd²

12f³ - 4f³ + 15f²d - 4f²d - 4f²d + f²d - 5fd² + fd² = 0

8f³ + 8f²d - 4fd² = 0

4f(2f² + 2fd - d²) = 0

2f² + 2fd - d² = 0

∴ d² - 2fd - 2f² = 0

The quadratic formula is given as:

d = (-b ± √(b² - 4ac))/(2a)

Where,

a = 1; b = -2f; c = -2f²

On substituting the values, we get,

d = (2f ± √(4f² + 8f²))/2

d = (2f ± √(12f²))/2

d = (2f ± 2f√(3f²))/2

∴ d = (f ± f√(3))

If d = f - f√3, we get,

4f - d = 4f - f + f√3 = 3f + f√3

This point lies beyond radius of curvature which is impossible.

If d = f + f√3, we get,

The point 'O' is located at a distance of f + f√3 from the convex mirror.