A beaker contains water (refractive index \( \mu_{w} \)) up to a height \( h \) and kerosene oil (refractive index \( \mu_{o} \)) above water up to another height \( H \). What will be the apparent shift in the position of an ink dot marked on the bottom of the beaker when viewed from above?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( (1 - 1/\mu_{w})h + (1 - 1/\mu_{o})H \)
A layer of thickness t and refractive index \( \mu \) makes an object below it appear raised by the normal shift \( t(1 - 1/\mu) \), for near-normal viewing. The shifts of successive layers simply add.
\( \displaystyle \text{shift due to water} = h\left(1 - \frac{1}{\mu_{w}}\right) \)
\( \displaystyle \text{shift due to kerosene} = H\left(1 - \frac{1}{\mu_{o}}\right) \)
\( \displaystyle \text{total shift} = \left(1 - \frac{1}{\mu_{w}}\right)h + \left(1 - \frac{1}{\mu_{o}}\right)H \)
Each factor must go with the thickness of its own liquid: water (\( \mu_{w} \)) fills height h and kerosene (\( \mu_{o} \)) fills height H.