Question

a covex lens of f=20cm has a object placed on its principal axis 40 cm away from it. a plane mirror is kept 30 cm behind the lens. what is the position of the image?

Answer 1

The position of the image is 10 cm behind the lens.

Answer 2

Explanation of the solution:

1. Image formation by the convex lens (first pass):

• The object (O) is placed at a distance $u_1 = -40$ cm from the convex lens (using Cartesian sign convention, where the object is on the left and light travels from left to right).
• The focal length of the convex lens is $f = +20$ cm.
• Using the lens formula:

$\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f}$

$\frac{1}{v_1} - \frac{1}{(-40)} = \frac{1}{20}$

$\frac{1}{v_1} + \frac{1}{40} = \frac{1}{20}$

$\frac{1}{v_1} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40} = \frac{1}{40}$

$v_1 = +40$ cm

• This means the first image ($I_1$) is formed 40 cm to the right of the lens. Since $v_1$ is positive, it is a real image.

2. Image formation by the plane mirror:

• The plane mirror (M) is kept 30 cm behind the lens.
• The image $I_1$ is formed 40 cm behind the lens.
• Since $I_1$ is 40 cm from the lens and the mirror is 30 cm from the lens, $I_1$ is located behind the plane mirror.
• The distance of $I_1$ from the mirror is $40 \text{ cm} - 30 \text{ cm} = 10 \text{ cm}$.
• When an object is behind a plane mirror, it acts as a virtual object for the mirror. A plane mirror forms a real image of a virtual object at the same distance in front of the mirror.
• So, the image formed by the mirror ($I_2$) will be 10 cm in front of the mirror.
• The position of $I_2$ relative to the lens is (Distance of mirror from lens) - (Distance of $I_2$ from mirror) = $30 \text{ cm} - 10 \text{ cm} = 20 \text{ cm}$.
• Thus, $I_2$ is formed 20 cm to the right of the lens.

3. Image formation by the convex lens (second pass):

• The light rays reflected from the mirror travel back towards the lens. The image $I_2$ (located 20 cm to the right of the lens) acts as an object for the lens for this second pass.
• Since $I_2$ is a real object located to the right of the lens, and light is now incident from the right, the object distance for the lens is $u_2 = +20$ cm (using the convention that real objects on the side from which light is incident have positive 'u' when light is incident from that side, or simply considering distances from the lens, $I_2$ is at $+20$ cm).
• The focal length of the lens is $f = +20$ cm.
• Using the lens formula again:

$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f}$

$\frac{1}{v_2} - \frac{1}{(+20)} = \frac{1}{20}$

$\frac{1}{v_2} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$

$v_2 = +10$ cm

• This means the final image ($I_3$) is formed 10 cm to the right of the lens. Since $v_2$ is positive, it is a real image.

The final image is formed 10 cm behind the lens.