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A microscope is focused on a mark on a piece of paper and then a slab of glass of thickness $3\,cm$ and a refractive index of 1.5 is placed over the mar. How should the microscope be moved to get the mark in focus again?
A) $1\,cm$ upward
B) $4.5\,cm$ downward
C) $1\,cm$ downward
D) $2\,cm$ upward

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Last updated date: 13th Jun 2024
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Answer
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Hint:We know that the refractive index of a medium is given by the ratio of real depth to apparent depth. Using this we can find the apparent depth. The distance to which microscope should be moved to get the new mark in focus will be equal to the distance to which the mark is raised when the glass slab is placed.

Complete step by step solution:
It is given that a microscope is focused on a mark.
A glass slab of $3\,cm$ thickness and refractive index 1.5 is placed over a mark. We need to find how the microscope should be moved to focus the mark again.
We know that when a glass rod is placed then the mark will appear at an apparent depth which is less than the real depth.
The refractive index of a medium is given by the ratio of real depth to apparent depth.
That is,
$\mu = \dfrac{{{\text{real}}\,{\text{depth}}}}{{{\text{apparent}}\,{\text{depth}}}}$
Using this we get apparent depth as the ratio of real depth to refractive index.
${\text{apparent}}\,{\text{depth}} = \dfrac{{{\text{real}}\,{\text{depth}}}}{\mu }$
Since the height of the slab is given as $3\,cm$ the real depth can be taken as $3\,cm$ .
By substituting the values we get the apparent depth as
${\text{apparent}}\,{\text{depth}} = \dfrac{3}{{1.5}}$
$\therefore {\text{apparent}}\,{\text{depth}} = 2\,cm$
This is the apparent depth from the top of the glass slab. We need to find how much the mark is raised .
That is $3\,cm - 2\,cm = 1\,cm$ . This is the distance to which the mark is raised.
The focal length of a microscope is the distance between the lens and the surface of the object when a clear image of the object can be seen. Suppose the focal length is $x$ initially.
When the object is raised this distance get reduced to $x - 1$,
The image will be at focus only if the length is $x$ so we need to move the microscope upward through a distance $1\,cm$ so that the object will be again at focus.

So, the correct answer is option A.

Note: Remember that the apparent depth in a denser medium will be less than the real depth. So, the point will appear raised in a denser medium. This apparent depth is the distance from the top of the glass slab to the place where the image appears. We need the distance the mark is raised so subtract the apparent depth from the real depth to get the distance to which mark is raised.