Question

A man observes a coin placed at the bottom of a beaker which contains two immiscible liquids of refractive indices \[1.2\] and \[1.4\] as shown in the figure. A plane mirror is also placed on the surface of liquid. The distance of image (from mirror) of coin in mirror as seen from medium \[1.2\] by an observer just above the boundary of the two media is

A. 18 cm
B. 12 cm
C. 9 cm
D. none of these

Answer 1

Hint: The ratio of real depth and apparent depth of an object give us the refractive index of the medium. Using this equation we can find apparent depth of the object in two mediums with different refractive indices. But the image distance is twice the real distance. Once the image distance of the coin is obtained, we can find out the distance of image of coin in mirror seen from 2nd medium.

Formula used:
\[\text{Apparent depth= }\dfrac{\text{Real depth}}{\text{Refractive index of the medium}}\]

Complete answer:

Height of liquid with refractive index \[1.2\] \[=3cm\]
Height of liquid with refractive index \[1.4\] \[=7cm\]
We have,
Refractive index of a medium, \[\text{ }\!\!\mu\!\!\text{ = }\dfrac{\text{Real depth}}{\text{Apparent depth}}\]
Then,
\[\text{Apparent depth= }\dfrac{\text{Real depth}}{\text{Refractive index of the medium}}\] ---------- 1
Apparent depth while viewing from the interface of two liquids,
\[{{\text{h}}_{1}}\text{= }\dfrac{7}{\text{1}\text{.4}}=5cm\]
Apparent depth while viewing from the interface of two mediums,
\[{{\text{h}}_{2}}\text{= }\dfrac{3}{\text{1}\text{.2}}=2.5cm\]
Total apparent depth, \[\text{h = }{{\text{h}}_{1}}+{{h}_{2}}=5+2.5=7.5cm\]
Image distance of coin \[=7.5\times 2=15cm\]
For an observer in 2nd medium, the distance of image of coin in mirror seen from medium with 1.2 refractive index \[=15-3=12cm\]

So, the correct answer is “Option B”.

Note:
When light rays bound off a reflective surface, a mirror image is obtained. When we look into a mirror, we see an image of ourselves behind the glass. For a plane mirror, the distance of the image from us will be twice the real distance. For all real objects, a plane mirror produces only virtual images, which are erect and has the same size as the object.