Optics Question 124
Question: A beaker containing liquid is placed on a table, underneath a microscope which can be moved along a vertical scale. The microscope is focussed, through the liquid onto a mark on the table when the reading on the scale is a. It is next focussed on the upper surface of the liquid and the reading is b. More liquid is added and the observations are repeated, the corresponding readings are c and d. The refractive index of the liquid is
Options:
A) $ \frac{d-b}{d-c-b+a} $
B) $ \frac{b-d}{d-c-b+a} $
C) $ \frac{d-c-b+a}{d-b} $
D) $ \frac{d-b}{a+b-c-d} $
Show Answer
Answer:
Correct Answer: A
Solution:
The real depth $ =\mu $ ( apparent depth)
$ \Rightarrow $ In first case, the real depth $ h _{1}=\mu (b-a) $
Similarly in the second case, the real depth $ h _{2}=\mu (d-c) $
Since $ h _{2}>h _{1}, $ the difference of real depths $ =h _{2}-h _{1}=\mu (d-c-b+a) $
Since the liquid is added in second case, $ h _{2}-h _{1}=(d-b) $
$ \Rightarrow \mu =\frac{(d-b)}{(d-c-b+a)} $
