MHT CET · Physics · Ray Optics
When final image is formed at D.D.V from eye, the magnifying power of microscope is ( \(f\) is the focal length of the lens)
- A \(1+\frac{f}{D}\)
- B \(1+\frac{D}{f}\)
- C \(\frac{D}{f}\)
- D \(1-\frac{D}{f}\)
Answer & Solution
Correct Answer
(B) \(1+\frac{D}{f}\)
Step-by-step Solution
Detailed explanation
Consider the diagram below:
Given, \(v=(-D)\),
Using mirror formula: \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\begin{aligned} & \Rightarrow \frac{1}{(-D)}-\frac{1}{u}=\frac{1}{f} \\ & \Rightarrow \frac{1}{u}=-\left(\frac{f+D}{f D}\right)\end{aligned}\)
Now, magnification can be easily calculated as:
\(\therefore m=\frac{v}{u}=\frac{(-D)(f+D)}{(-f D)}=\left(1+\frac{D}{f}\right)\)
Given, \(v=(-D)\),
Using mirror formula: \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\begin{aligned} & \Rightarrow \frac{1}{(-D)}-\frac{1}{u}=\frac{1}{f} \\ & \Rightarrow \frac{1}{u}=-\left(\frac{f+D}{f D}\right)\end{aligned}\)
Now, magnification can be easily calculated as:
\(\therefore m=\frac{v}{u}=\frac{(-D)(f+D)}{(-f D)}=\left(1+\frac{D}{f}\right)\)
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