Question

One face of glass cube of refractive index 1.5 and
side 6 cm is silvered. An object is placed at a
distance 7 cm from the face opposite to the
silvered face. Image of object behind the silvered
face is at
(1) 18 cm
(2) 9 cm
(3) 10 cm
(4) 8 cm​

Answer 1

Answer:

Final image is at distance 9 cm from the silvered surface

Explanation:

first image will form due to refraction of light at first surface

so we will have

$\frac{\mu_2}{d_i} = \frac{\mu_1}{d_o}$

$\frac{1.5}{d_i} = \frac{1}{7}$

so we have

$d_i = 10.5 cm$

now distance of the image for 2nd surface is given as

$d_o = 10.5 + 6$

$d_o = 16.5 cm$

now the reflection will occur at 2nd surface so we have image distance given as

$d_i = d_o = 16.5 cm$

now again refraction from the first surface

$\frac{1.5}{16.5 + 6} = \frac{1}{d_i}$

$d_i = 15 cm$

so distance from silvered surface is given as

$d = 15 - 6 = 9 cm$

#Learn

Topic : Reflection of light

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