
The lens of a simple magnifier has a focal length of 2.5 cm. Calculate the angular magnification produced when (a) the image is at least distance of distinct vision, and (b) infinity.
(A) 11, 10
(B) 11, 13
(C) 10, 13
(D) 11, 12
Answer
233.4k+ views
Hint Use the lens formula relating focal length f, image distance v and object distance u. Multiply by v on both sides to get the expression of magnification. Now, substitute the data in the expression to find the magnification.
Complete step-by-step answer
As we know that lens formula is given as
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
Multiplying with v on both sides
\[
\Rightarrow 1 - \dfrac{v}{u} = \dfrac{v}{f} \\
\Rightarrow 1 - {\text{ }}m{\text{ }} = \dfrac{v}{f} \\
\Rightarrow 1 - {\text{ }}\dfrac{v}{f} = m \\
\]
This gives the linear magnification of the lens.
Angular magnification is given as:
\[{m_\alpha }{\text{ }} = {\text{ }}1{\text{ }} + \dfrac{v}{f}\]
Substituting the values of v and f
\[
v{\text{ }} = {\text{ }}2.5cm \\
f{\text{ }} = {\text{ }}25{\text{ }}cm \\
\]
\[{m_\alpha }{\text{ }} = {\text{ }}1 + {\text{ }}\dfrac{{25}}{{2.5}}\]
\[{m_\alpha }{\text{ }} = {\text{ }}1 + {\text{ }}10\]
\[{m_\alpha }{\text{ }} = {\text{ }}11\]
When the object is placed at infinity, its angular magnification is given as:
\[{m_\alpha }{\text{ }} = {\text{ }}\dfrac{v}{f}\]
\[{m_\alpha }{\text{ }} = {\text{ }}\dfrac{{25}}{{2.5}}\]
\[{m_\alpha }{\text{ }} = {\text{ }}10\]
Therefore the correct answer is option A
Note There is a difference between angular magnification and linear magnification. Linear magnification is simply the ratio of image distance to object distance, whereas angular magnification is a derived formula. Magnification is used in optical instruments like microscope and telescope.
Complete step-by-step answer
As we know that lens formula is given as
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
Multiplying with v on both sides
\[
\Rightarrow 1 - \dfrac{v}{u} = \dfrac{v}{f} \\
\Rightarrow 1 - {\text{ }}m{\text{ }} = \dfrac{v}{f} \\
\Rightarrow 1 - {\text{ }}\dfrac{v}{f} = m \\
\]
This gives the linear magnification of the lens.
Angular magnification is given as:
\[{m_\alpha }{\text{ }} = {\text{ }}1{\text{ }} + \dfrac{v}{f}\]
Substituting the values of v and f
\[
v{\text{ }} = {\text{ }}2.5cm \\
f{\text{ }} = {\text{ }}25{\text{ }}cm \\
\]
\[{m_\alpha }{\text{ }} = {\text{ }}1 + {\text{ }}\dfrac{{25}}{{2.5}}\]
\[{m_\alpha }{\text{ }} = {\text{ }}1 + {\text{ }}10\]
\[{m_\alpha }{\text{ }} = {\text{ }}11\]
When the object is placed at infinity, its angular magnification is given as:
\[{m_\alpha }{\text{ }} = {\text{ }}\dfrac{v}{f}\]
\[{m_\alpha }{\text{ }} = {\text{ }}\dfrac{{25}}{{2.5}}\]
\[{m_\alpha }{\text{ }} = {\text{ }}10\]
Therefore the correct answer is option A
Note There is a difference between angular magnification and linear magnification. Linear magnification is simply the ratio of image distance to object distance, whereas angular magnification is a derived formula. Magnification is used in optical instruments like microscope and telescope.
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