Answer
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Hint: We have to find proper relations with the focal length of the magnifying glass and the magnification of the lens. We can use appropriate relations between the different parameters like image distance and object distance to relate them.
Complete step by step solution:
We are given a magnifying glass with focal length 2.5cm. The magnifying glasses that we usually use are the bifocal convex lenses. We are given the focal length of a convex lens for which we need to find the magnification possible.
We know that the lenses equation for a convex lens as –
\[\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\]
From this equation we can find the relation between the image distance, v and the object distance, u as the focal length of the lens is given as 2.5cm. Also, we know that these two quantities – the image distance and the object distance are directly related to the magnification of an image formed.
But we are given the focal length in order to find the magnifying power of magnifying glass. The magnifying power is found by relating it to the least distance of distinct vision (LDDV) and the power of the lens used. The relation is given as –
\[m=(LDDV\times Power)+1\]
We know that the power of the lens is defined as the inverse of the focal length in meters. So, we can find the power of the magnifying glass as –
\[\begin{align}
& \text{Power, }P=\dfrac{1}{f} \\
& \therefore P=\dfrac{1}{0.025}=40D \\
\end{align}\]
Now, we can find the magnification by substituting the power in the formula for magnification as –
\[\begin{align}
& \text{LDDV}=25cm=0.25m \\
& m=0.25\times 40+1 \\
& m=10+1 \\
& \therefore m=11times \\
\end{align}\]
The required magnification is given as 11 times.
The correct answer is B.
Note:
The magnification is considered such that the image is formed at the least distance of distinct vision in this case. But the higher magnifying equipment are made such as have an image at infinity for a much relation of the human eyes while observing the image.
Complete step by step solution:
We are given a magnifying glass with focal length 2.5cm. The magnifying glasses that we usually use are the bifocal convex lenses. We are given the focal length of a convex lens for which we need to find the magnification possible.
We know that the lenses equation for a convex lens as –
\[\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\]
From this equation we can find the relation between the image distance, v and the object distance, u as the focal length of the lens is given as 2.5cm. Also, we know that these two quantities – the image distance and the object distance are directly related to the magnification of an image formed.
But we are given the focal length in order to find the magnifying power of magnifying glass. The magnifying power is found by relating it to the least distance of distinct vision (LDDV) and the power of the lens used. The relation is given as –
\[m=(LDDV\times Power)+1\]
We know that the power of the lens is defined as the inverse of the focal length in meters. So, we can find the power of the magnifying glass as –
\[\begin{align}
& \text{Power, }P=\dfrac{1}{f} \\
& \therefore P=\dfrac{1}{0.025}=40D \\
\end{align}\]
Now, we can find the magnification by substituting the power in the formula for magnification as –
\[\begin{align}
& \text{LDDV}=25cm=0.25m \\
& m=0.25\times 40+1 \\
& m=10+1 \\
& \therefore m=11times \\
\end{align}\]
The required magnification is given as 11 times.
The correct answer is B.
Note:
The magnification is considered such that the image is formed at the least distance of distinct vision in this case. But the higher magnifying equipment are made such as have an image at infinity for a much relation of the human eyes while observing the image.
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