Question

A glass slab of tinkness 3cm and refractive index in place in front of concave mirror of f=20cm. where should point object is placed it into

Answer 1

41 cm

Answer 2

To form an image on itself with a concave mirror, the object must be placed at the center of curvature of the mirror.
The focal length of the concave mirror is given as $f = 20 \text{ cm}$.
The radius of curvature $R$ for a concave mirror is $R = 2f$.
So, $R = 2 \times 20 \text{ cm} = 40 \text{ cm}$.

A glass slab of thickness $t = 3 \text{ cm}$ and refractive index $\mu$ is placed in front of the mirror. From the similar question, we infer $\mu = 1.5$.

When light from the object passes through the glass slab before reaching the mirror, the apparent position of the object shifts. The shift $s$ is given by the formula: $s = t \left(1 - \frac{1}{\mu}\right)$

Substitute the given values: $s = 3 \text{ cm} \left(1 - \frac{1}{1.5}\right)$ $s = 3 \text{ cm} \left(1 - \frac{2}{3}\right)$ $s = 3 \text{ cm} \left(\frac{3-2}{3}\right)$ $s = 3 \text{ cm} \left(\frac{1}{3}\right)$ $s = 1 \text{ cm}$

This shift means that the object, as seen from the mirror, appears to be $1 \text{ cm}$ closer to the mirror than its actual position.
Let the actual distance of the object from the mirror be $u$.
The apparent distance of the object from the mirror, as perceived by the mirror, is $u_{apparent} = u - s$.

For the image to form on the object itself, the rays from the object must strike the mirror normally. For a concave mirror, this occurs when the object is placed at the center of curvature. Therefore, the apparent position of the object must coincide with the center of curvature. $u_{apparent} = R$ So, $u - s = R$

Substitute the values of $s$ and $R$: $u - 1 \text{ cm} = 40 \text{ cm}$ $u = 40 \text{ cm} + 1 \text{ cm}$ $u = 41 \text{ cm}$

Thus, the point object should be placed at a distance of 41 cm from the concave mirror.