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# A convex lens produces a real image m times the size of the object. What will be the distance of the object from the lens?$\text{A}\text{. }(m-1)f$$\text{B}\text{. }\left( \dfrac{m+1}{m} \right)f$$\text{C}\text{. }\left( \dfrac{m-1}{m} \right)f$$\text{D}\text{. }\left( \dfrac{m+1}{f} \right)$

Last updated date: 20th Jun 2024
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Hint: Use the formula for magnification due to a lens. With this find an expression for the position of the image. Then use the lens formula and substitute the value of the position of image. Hence, find the expression for the object position from the lens.

Formula used: magnification $=\dfrac{v}{u}$
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

It is given that a convex lens produces a real image. The size of the image produced is said to m times larger than the size of the object.
We will use the formula for magnification. The magnification is the ratio of the height of the image to the height of the image. A real image is always inverted. Therefore, the value of magnification is negative. This means that magnification is (-m).
The value of magnification is equal to $\dfrac{v}{u}$, where v and u are the distances of the image and the object from the lens.
$\Rightarrow -m=\dfrac{v}{u}$
$\Rightarrow v=-mu$ …. (i).
The relation between the position of the object and its image is given as $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ …. (ii), where f is the focal length of the lens.
Substitute the value of v from equation (i) into equation (ii).
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{-mu}-\dfrac{1}{u}$
$\Rightarrow \dfrac{1}{f}=\left( \dfrac{-1}{m}-1 \right)\dfrac{1}{u}$
$\Rightarrow \dfrac{1}{f}=\left( \dfrac{-1-m}{m} \right)\dfrac{1}{u}$
$\Rightarrow u=\left( \dfrac{-1-m}{m} \right)f$.
$\Rightarrow u=-\left( \dfrac{m+1}{m} \right)f$
This expression is for the position of the object but distance is a positive value.
Therefore, the object distance is $\left( \dfrac{m+1}{m} \right)f$.

So, the correct answer is “Option B”.

Note: Note a real image formed by a real object is always inverted. Therefore, in this case the height of the image is negative and the height of the object is positive.
Since magnification is defined as the ratio of the height of the image to the height of the object, the value of magnification is negative.
When a real image is formed, the position of the image is positive and the position of the object is negative. Therefore, the ratio $\dfrac{v}{u}$ is negative. Again we prove that the magnification is negative.