Question

A convex and concave mirror, each of focal length fare placed at a distance 4f apart as shown in figure
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

Answer

The point source of light should be placed at a distance of (f + f√3) from convex mirror or (3f - f√3) from concave mirror between the two mirrors.

Explanation

(Refer attachment)

Let the point source of light be placed at O situated at a distance of 'd' from the convex mirror.

Now, the point source at 'O' will act as object for convex mirror and form image beyond the mirror, say at position Q

Now, for concave mirror, the ray appears to come from 'Q', hence 'Q' will act as object for concave mirror. The concave mirror will form its image at O.

So we get the following points

• O is object for convex mirror when Q is the image.
• O is object for convex mirror when Q is the image. Q is object for concave mirror when O is the image.

Now, we will apply mirror formula to find 'd'

Sign convention is important here as wrong sign convention will give wrong answer. We will follow the Cartesian rule. The focal length of convex mirror is positive and focal length of concave mirror is negative.

Mirror formula :

$\sf \frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

For convex mirror,

f = f

u = -d (object distance is always negative)

Let the image formed at Q be at a distance of v' from the pole of convex mirror

$\sf\frac{1}{f} = \frac{-1}{d} + \frac{1}{v'}$

$\sf\frac{1}{v'} = \frac{1}{d} + \frac{1}{f}$

$\sf\frac{1}{v'} = \frac{f +d}{fd}$

$\sf v' = \frac{fd}{f +d}$

Now, Q acts as object for concave mirror and O as image for the same.

So, for concave mirror,

f = - f

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

→ u = -(v' + 4f)

→ u = $\sf -( \frac{fd}{f + d} + 4f)$

→ u = $\sf -(\frac{4f^2 + 5fd}{f + d})$

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

→ v = - (4f - d)

So,

→ $\sf\frac{-1}{f} = \frac{-(f + d)}{4f^2 + 5fd} - \frac{1}{4f -d}$

Cancel negative sign from both sides

→ $\sf\frac{1}{f} = \frac{f+d}{4f^2 + 5fd} + \frac{1}{4f -d}$

→ $\sf\frac{f+d}{4f^2+5fd} =\frac{1}{f} - \frac{1}{4f-d}$

→ $\sf\frac{f+d}{4f^2+5fd} = \frac{3f-d}{4f^2-fd}$

Now cross multiply,

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

Put all terms on one side

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

Simplify

→ 8f³ + 8f²d - 4fd² = 0

→ 4f(2f² + 2fd - d²) = 0

→ 2f² + 2fd - d² = 0

→ d² - 2fd - 2f² = 0

Apply quadratic formula

$\sf d = \frac{2f\pm \sqrt{4f^2 + 8f^2}}{2}$

$\sf d = \frac{2f\pm\sqrt{12f^2}}{2}$

$\sf d = \frac{2f\pm2f\sqrt{3}}{2}$

$\sf d = f\pm f\sqrt{3}$

Now, consider the concave mirror. 'Q' is beyond the centre of curvature. (centre of curvature = 2f). So if it's beyond the centre of curvature, the image should be formed between centre of curvature and focus. So, image should be formed between 2f and f.

Now, if we take d as f - f√3, then

4f - d = 4f - f + f√3 = 3f + f√3

3f + f√3 lies beyond the centre of curvature, hence this value of d is wrong.

So, value of d = f + f√3.

Hence, O is situated at a distance of f + f√3 from the convex mirror.