
In the figure shown, find the total magnification after two successive reflections first on ${M_1}$ and then on ${M_2}$ .

A. $ + 1$
B. $ - 2$
C. $ + 2$
D. $ - 1$
Answer
160.8k+ views
Hint: The two reflections M1 and M2's magnification must first be determined. Here, we need to use the mirror formula to find the image distance i.e, v. The value of u, is already given. After that, we can find the total magnification.
Formula used:
1. Mirror Formula:
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ … (1)
Here, $f$ is the focal length of the mirror,
$u$ is the object distance and
$v$ is the image distance.
Magnification :
$M=-\dfrac{v}{u}$
Magnification = ${M_1} \times {M_2}$
Complete answer:
Let’s first determine $M_{1}$’s magnification:
Given:
Mirror1’s distance from the object is u=-30
The focal length is f=-20cm
The image’s distance from the mirror is given by, v which needs to be determined.
According to the mirror formula,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Rearranging the equation,
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$
By putting the values in the above equation we get that,
$\dfrac{1}{v} = \dfrac{1}{{ - 20}} + \dfrac{1}{{30}}$
Further solving we get,
$v = \dfrac{{ - 20 \times 30}}{{10}}$
By completing the calculation we get the value of v as,
$v = - 60cm$
Now $M_{1}$ will be,
$M_{1}=-\dfrac{v}{u}$
By putting the values in the above equation we get,
$M_{1}=-\dfrac{-60}{-30}$
$M_{1}=-2$
Next, determine $M_{2}$'s magnification:
Given:
Mirror1’s distance from the object is u=20 (The image that is created will serve as an object for ${M_2}$ and will be located 20cm left of ${M_2}$ )
The focal length is f=10cm
The image’s distance from the mirror is given by, v which needs to be determined.
Let’s put values in the above-used mirror formula to find v.
$\dfrac{1}{v} = \dfrac{1}{{10}} - \dfrac{1}{{20}}$
Further solving we get,
$v = \dfrac{{10 \times 20}}{{10}} = 20cm$
Now $M_{2}$ will be,
$M_{2}=-\dfrac{v}{u}$
By putting the values in the above equation we get,
$M_{2}=-\dfrac{20}{20}$
$M_{1}=-1$
The total magnification can be determined by the below-mentioned formula,
Magnification = ${M_1} \times {M_2}$
By putting the value we get that,
Magnification = $M=\lgroup-2\rgroup\times\lgroup-1\rgroup$
Therefore, the total magnification is 2.
The correct option is C.
Note:Be careful regarding sign convention. All distances are calculated starting at the mirror's pole. Also, the image that is created will serve as an object for ${M_2}$. So, don’t take the object distance for ${M_2}$ to be 30 cm by blindly following the diagram.
Formula used:
1. Mirror Formula:
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ … (1)
Here, $f$ is the focal length of the mirror,
$u$ is the object distance and
$v$ is the image distance.
Magnification :
$M=-\dfrac{v}{u}$
Magnification = ${M_1} \times {M_2}$
Complete answer:
Let’s first determine $M_{1}$’s magnification:
Given:
Mirror1’s distance from the object is u=-30
The focal length is f=-20cm
The image’s distance from the mirror is given by, v which needs to be determined.
According to the mirror formula,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Rearranging the equation,
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$
By putting the values in the above equation we get that,
$\dfrac{1}{v} = \dfrac{1}{{ - 20}} + \dfrac{1}{{30}}$
Further solving we get,
$v = \dfrac{{ - 20 \times 30}}{{10}}$
By completing the calculation we get the value of v as,
$v = - 60cm$
Now $M_{1}$ will be,
$M_{1}=-\dfrac{v}{u}$
By putting the values in the above equation we get,
$M_{1}=-\dfrac{-60}{-30}$
$M_{1}=-2$
Next, determine $M_{2}$'s magnification:
Given:
Mirror1’s distance from the object is u=20 (The image that is created will serve as an object for ${M_2}$ and will be located 20cm left of ${M_2}$ )
The focal length is f=10cm
The image’s distance from the mirror is given by, v which needs to be determined.
Let’s put values in the above-used mirror formula to find v.
$\dfrac{1}{v} = \dfrac{1}{{10}} - \dfrac{1}{{20}}$
Further solving we get,
$v = \dfrac{{10 \times 20}}{{10}} = 20cm$
Now $M_{2}$ will be,
$M_{2}=-\dfrac{v}{u}$
By putting the values in the above equation we get,
$M_{2}=-\dfrac{20}{20}$
$M_{1}=-1$
The total magnification can be determined by the below-mentioned formula,
Magnification = ${M_1} \times {M_2}$
By putting the value we get that,
Magnification = $M=\lgroup-2\rgroup\times\lgroup-1\rgroup$
Therefore, the total magnification is 2.
The correct option is C.
Note:Be careful regarding sign convention. All distances are calculated starting at the mirror's pole. Also, the image that is created will serve as an object for ${M_2}$. So, don’t take the object distance for ${M_2}$ to be 30 cm by blindly following the diagram.
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