Question
A plane mirror is held at a height $$h$$ above the bottom of an empty beaker. The beaker is now filled with water up to depth $$d.$$ The general expression for the distance from a scratch at the bottom of the beaker to its image in terms of $$h$$ and the depth $$d$$ of water in the beaker is
A plane mirror is held at a height $$h$$ above the bottom of an empty beaker. The beaker is now filled with water up to depth $$d.$$ The general expression for the distance from a scratch at the bottom of the beaker to its image in terms of $$h$$ and the depth $$d$$ of water in the beaker is
A.
$$2h - d\left( {\frac{\mu }{{\mu - 1}}} \right)$$
B.
$$2h - \frac{d}{2}\left( {\frac{{\mu - 1}}{\mu }} \right)$$
C.
$$2h - d\left( {\frac{{\mu - 1}}{\mu }} \right)$$
D.
$$2h - d\left( {\frac{{2\mu - 1}}{\mu }} \right)$$
Answer :
$$2h - d\left( {\frac{{\mu - 1}}{\mu }} \right)$$
Solution :
The distance of bottom of the beaker from mirror $$ = h - d\left( {1 - \frac{1}{\mu }} \right)$$
So it will be at a distance $$ = h - d\left( {1 - \frac{1}{\mu }} \right)$$ from mirror. Now distance between bottom of beaker and image
$$\eqalign{ & = h + h - d\left( {1 - \frac{1}{\mu }} \right) \cr & = 2h - d\left( {\frac{{\mu - 1}}{\mu }} \right). \cr} $$
The distance of bottom of the beaker from mirror $$ = h - d\left( {1 - \frac{1}{\mu }} \right)$$
So it will be at a distance $$ = h - d\left( {1 - \frac{1}{\mu }} \right)$$ from mirror. Now distance between bottom of beaker and image
$$\eqalign{ & = h + h - d\left( {1 - \frac{1}{\mu }} \right) \cr & = 2h - d\left( {\frac{{\mu - 1}}{\mu }} \right). \cr} $$
