First slide

Refraction of light

Question

An object AB of height 2 cm is placed at distance 63 cm from a concave mirror of radius of curvature 40 cm. A slab of refractive index $\frac{4}{3}$ and thickness 12 cm is placed between the object and the mirror.

Difficult

A

Final image is real, inverted and 2 cm high
B

Final image is virtual, inverted and 1 cm
C

Final image is 33 cm from the pole of mirror
D

Final image is 27 cm from the pole of mirror

Solution

The object shift due to slab is $x = t (1 - \frac{1}{u}) = 12 (1 - \frac{3}{4}) = 3 c m$

The image $A_{1} B_{1}$ formed due to refraction through slab behaves as object for concave mirror.
For concave mirror, $u = - 60 c m, f = - 20 c m$
According to mirror formula,
$∵ \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{v} - \frac{1}{60} = - \frac{1}{20}$
Or $\frac{1}{v} = \frac{1}{60} - \frac{1}{20} = \frac{1 - 3}{60}$
$∴ v = - 30 c m$
$m = \frac{A_{1} B_{2}}{A_{1} B_{1}} = - \frac{v}{u} = - \frac{(- 30)}{- 60}$
$∴ A_{2} B_{2} = - 1 c m$
$A_{2} B_{2}$ behaves as virtual object for second times refraction through slab. The final image is shifted 3cm leftward from $A_{2} B_{2}$ due to second times refraction through slab. Thus, final image is real, 1 cm high inverted and 33 cm from the concave mirror. The ray diagram is shown in the figure.

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