
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. $2h - d\left( {\dfrac{\mu }{{\mu - 1}}} \right)$
B. $2h - \dfrac{d}{2}\left( {\dfrac{{\mu - 1}}{\mu }} \right)$
C. $2h - d\left( {\dfrac{{\mu - 1}}{\mu }} \right)$
D. $2h - d\left( {\dfrac{{2\mu - 1}}{\mu }} \right)$
Answer
164.1k+ views
Hint: Generally for a plane mirror, the distance of the image from us will be two times of the real distance as the image is the same distance behind the mirror as the object is in front of the mirror. First we must find out the distance between image and mirror by using apparent shift. Then we have to add the distance of the scratch and distance of the image from the mirror to get the final answer.
Formula used:
Shift in location of image/object due to a slab:
$d\left( {1 - \dfrac{1}{\mu }} \right)$.
Complete answer:
In the question, we have given that the height of the plane mirror is $h$ above the level of an empty beaker, And, the depth of the water in a beaker is $d$.
In order to know the equation for a plane mirror to reflect an image, at least two light rays from the object must either intersect or appear to intersect at the same location.
Then we have the shift observed is:
$d\left( {1 - \dfrac{1}{\mu }} \right)$.
The object distance with respect to the mirror is $h - shift$, so the distance of the image from the mirror is:
$h - d\left( {1 - \dfrac{1}{\mu }} \right)$,
As a result, the distance of an image from the scratch at the bottom of the beaker is as follows:
$h - d\left( {1 - \dfrac{1}{\mu }} \right) + h$
On expanding the above equation, then we obtain:
$2h - d\left( {1 - \dfrac{1}{\mu }} \right) \\$
$= 2h - d\left( {\dfrac{{\mu - 1}}{\mu }} \right) \\$
Thus, the correct option is (C) $2h - d\left( {\dfrac{{\mu - 1}}{\mu }} \right)$
Note: To approach these types of problems we must remember the formula such as apparent shift. Also students should remember the main characteristics of a plane mirror which will help them during their exam to deal with numerical problems for example the distance between the object and plane mirror is exactly same as the distance between mirror plane and image.
Formula used:
Shift in location of image/object due to a slab:
$d\left( {1 - \dfrac{1}{\mu }} \right)$.
Complete answer:
In the question, we have given that the height of the plane mirror is $h$ above the level of an empty beaker, And, the depth of the water in a beaker is $d$.
In order to know the equation for a plane mirror to reflect an image, at least two light rays from the object must either intersect or appear to intersect at the same location.
Then we have the shift observed is:
$d\left( {1 - \dfrac{1}{\mu }} \right)$.
The object distance with respect to the mirror is $h - shift$, so the distance of the image from the mirror is:
$h - d\left( {1 - \dfrac{1}{\mu }} \right)$,
As a result, the distance of an image from the scratch at the bottom of the beaker is as follows:
$h - d\left( {1 - \dfrac{1}{\mu }} \right) + h$
On expanding the above equation, then we obtain:
$2h - d\left( {1 - \dfrac{1}{\mu }} \right) \\$
$= 2h - d\left( {\dfrac{{\mu - 1}}{\mu }} \right) \\$
Thus, the correct option is (C) $2h - d\left( {\dfrac{{\mu - 1}}{\mu }} \right)$
Note: To approach these types of problems we must remember the formula such as apparent shift. Also students should remember the main characteristics of a plane mirror which will help them during their exam to deal with numerical problems for example the distance between the object and plane mirror is exactly same as the distance between mirror plane and image.
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