AP EAMCET · PHYSICS · Ray Optics
A lens forms real and virtual images of an object, when the object is at \(u_1\) and \(u_2\) distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as \(m\) )
- A \(\left(\frac{u_1+u_2}{2}\right) m\)
- B \(\left(\frac{u_1-u_2}{3}\right) 2 m\)
- C \(\left(\frac{u_1-u_2}{2}\right) 3 m\)
- D \(\left(\frac{u_1+u_2}{3}\right) 2 m\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{u_1-u_2}{2}\right) 3 m\)
Step-by-step Solution
Detailed explanation
Lens maker formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) Case 1 Real image \(v\) and \(f\) are positive, \(u\) is negative, so \( \frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f} \Rightarrow \frac{u_1}{v_1}+1=\frac{u_1}{f} \) Since, magnification for real image,…
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