Question

A plane mirror is made of glass slab (refractive indec of glas =1.5) 2.5 cm thick and silvered on back.A point source is placed 5cm front of the unsilvered face of the mirror.What will be the position of final image?

Answer 1

Answer on Question #73942, Physics / Optics

Question. A plane mirror is made of glass slab (refractive index of glas $n = 1.5$) $d = 2.5$ cm thick and silvered on back. A point source is placed $p = 5$ cm front of the unsilvered face of the mirror. What will be the position of final image?

Given. $n = 1.5$; $d = 2.5$ cm; $p = 5$ cm.

Find. $x - ?$

Solution.

Let

1) $I_{1}$ is the image formed by refraction from $P_{1}$;

2) $I_{2}$ is the image formed by reflection from $P_{2}$;

3) $I_{1}$ is the image formed by again refraction from $P_{1}$.

Location of the image $I_{1}$

$$\frac {1}{p} = - \frac {n}{q} \rightarrow q = - n \cdot p = - 1. 5 \cdot 5 = - 7. 5 \mathrm {c m} \text { from } P _ {1}$$

or

$$7. 5 + 2. 5 = 1 0 \mathrm {c m} \text { from } P _ {2}.$$

Location of the image $I_{2}$

$$1 0 \mathrm {c m} \text { behind mirror or } 1 0 + 2. 5 = 1 2. 5 \mathrm {c m} \text { from } P _ {1}.$$

Location of the image $I_{3}$

$$\frac {n}{1 2 . 5} = \frac {1}{x} \rightarrow x = \frac {1 2 . 5}{n} = \frac {1 2 . 5}{1 . 5} = 8. 3 3 \mathrm {c m} \text { from unsilvered face } P _ {1}.$$

Answer. $x = 8.33$ cm from unsilvered face $P_{1}$.

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