The focal length of a convex lens is \[10\,cm.\] the magnifying power when it is used as a magnifying glass to form the image at (i) near point and (ii) far point are:
A. \[2.5,3.5\]
B. \[3.5,4.5\]
C. \[4.5,3.5\]
D. \[3.5,2.5\]
Answer
543.6k+ views
Hint: To solve this question we have to know about magnification. We know that magnification, otherwise called proliferation proportion, is a property of a camera focal point which depicts how intently you've centered. In particular, amplification is the proportion between an item's size when projected on a camera sensor versus its size in reality.
Complete step by step answer:
We know that, for the near point, magnification, \[{m_1} = 1 + \dfrac{D}{f}\]
Now, for the far point, \[{m_2} = \dfrac{D}{f}\]
Thus, putting $D$ is equal to \[25\] and f is equal to \[10\]
We will get, \[{m_2} = \dfrac{{25}}{{10}} = 2.5\] and
\[{m_1} = 1 + \dfrac{{25}}{{10}} \\
\Rightarrow {m_1} = 2.5 + 1 \\
\therefore {m_1} = 3.5 \\ \]
So, the correct answer is option D.
Note:we also can say, the magnification of convex lens is a proportion between the picture tallness and item stature. An amplification of 2 demonstrates the picture is double the size of the article and an amplification of 1 shows a picture size being equivalent to the item size. On the off chance that the amplification is positive, the picture is upstanding contrasted with the item (virtual picture). We also can say, the ratio of the size of the image formed by refraction from the lens to the size of the object, is called linear magnification produced by the lens.
Complete step by step answer:
We know that, for the near point, magnification, \[{m_1} = 1 + \dfrac{D}{f}\]
Now, for the far point, \[{m_2} = \dfrac{D}{f}\]
Thus, putting $D$ is equal to \[25\] and f is equal to \[10\]
We will get, \[{m_2} = \dfrac{{25}}{{10}} = 2.5\] and
\[{m_1} = 1 + \dfrac{{25}}{{10}} \\
\Rightarrow {m_1} = 2.5 + 1 \\
\therefore {m_1} = 3.5 \\ \]
So, the correct answer is option D.
Note:we also can say, the magnification of convex lens is a proportion between the picture tallness and item stature. An amplification of 2 demonstrates the picture is double the size of the article and an amplification of 1 shows a picture size being equivalent to the item size. On the off chance that the amplification is positive, the picture is upstanding contrasted with the item (virtual picture). We also can say, the ratio of the size of the image formed by refraction from the lens to the size of the object, is called linear magnification produced by the lens.
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