Question

A convex lens is kept coaxially on the left side of a concave mirror at a distance of 5f from it where f is focal length of each of them. An object is kept at a distance of 2f on the left side of the convex lens. Draw the ray diagram showing the formation of image by this combination. Find the distance of the final image from the mirror.

Hint:

Lens first: image at \(2f\) on the right, \(3f\) from the mirror. Mirror then gives an image at \(1.5f\) in front of it. Rays then pass through the lens again.

Answer 1

Step 1: Set up:
Let the lens be \(L\) (focal length \(f\)) and the concave mirror be \(M\) (focal length \(f\)), a distance \(5f\) apart. The object \(O\) is \(2f\) to the left of the lens. Take light travelling to the right as positive.

Step 2: Image by the lens:
\(u = -2f\). Lens formula: \(\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{f} - \dfrac{1}{2f} = \dfrac{1}{2f}\), so \(v = +2f\).
The first image \(I_1\) is real, inverted and equal in size, \(2f\) to the right of the lens. Its distance from the mirror is \(5f - 2f = 3f\).

Step 3: Image by the mirror:
For \(M\), \(I_1\) acts as the object at \(u = -3f\). The focal length of a concave mirror is \(F = -f\) in this convention.
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{F} \Rightarrow \frac{1}{v} = -\frac{1}{f} + \frac{1}{3f} = -\frac{2}{3f} \]
So \(v = -\dfrac{3f}{2}\). The mirror forms a real image \(I_2\) at \(1.5f\) in front of the mirror, between the mirror and the lens. Its magnification is \(-v/u = -1/2\), so it is diminished.

Step 4: Ray diagram (how to draw it):
1. Draw the axis, the lens at \(L\), the mirror at \(M\) (\(5f\) to the right of \(L\)), and mark \(f\) and \(2f\) on both sides of the lens.
2. Place the arrow object at \(2f\) left of the lens. Draw a ray parallel to the axis, which bends through the focus on the right, and a ray through the centre of the lens, which goes straight.
3. These two rays meet at \(2f\) on the right of the lens. This point is \(I_1\), \(3f\) from the mirror.
4. Beyond \(I_1\) the rays diverge and fall on the mirror. Draw one ray from \(I_1\) parallel to the axis, which reflects through the mirror focus, and one through the pole, which reflects symmetrically.
5. The reflected rays meet at \(I_2\), \(1.5f\) in front of the mirror.

Step 5: The returning rays and the lens again:
The reflected rays cross at \(I_2\) and then carry on to the left towards the lens. Distance of \(I_2\) from the lens is \(5f - 1.5f = 3.5f\), so \(I_2\) acts as a real object for the lens for rays that now move to the left.
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{3.5f} = \frac{5}{7f} \Rightarrow v = \frac{7f}{5} = 1.4f \]
A final real image \(I_3\) forms \(1.4f\) to the left of the lens. Its distance from the mirror is \(5f + 1.4f = 6.4f\).

Final Answer:
The mirror forms its image \(I_2\) at \(3f/2\) from the mirror (real, in front of it). If the reflected rays are followed back through the lens, the last image is at \(7f/5\) to the left of the lens, which is \(32f/5 = 6.4f\) from the mirror. \[ \boxed{\text{Image by the mirror: } \tfrac{3f}{2}\text{ from the mirror;\ after the lens again: } \tfrac{32f}{5}\text{ from the mirror}} \]