Q.42. A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cu

Q: What is the correct answer for this exam Physics Q42?

The correct answer is A. Sol. Magnification $M = \frac{L D}{F_o F_e}$. When the objective lens is cut vertically, its focal length changes. For a biconvex lens, $\frac{1}{f} = (n-1)(\frac{1}{R} + \frac{1}{R}) = (n-1)\frac{2}{R}$. When cut vertically, it becomes a plano-convex lens. For a plano-convex lens, $\frac{1}{f'} = (n-1)(\frac{1}{R} - \frac{1}{\infty}) = (n-1)\frac{1}{R}$. So, $f' = 2f$. To retain the same magnific

Q: Q.42. A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original…

The correct answer is A. Sol. Magnification $M = \frac{L D}{F_o F_e}$. When the objective lens is cut vertically, its focal length changes. For a biconvex lens, $\frac{1}{f} = (n-1)(\frac{1}{R} + \frac{1}{R}) = (n-1)\frac{2}{R}$. When cut vertically, it becomes a plano-convex lens. For a plano-convex lens, $\frac{1}{f'} = (n-1)(\frac{1}{R} - \frac{1}{\infty}) = (n-1)\frac{1}{R}$. So, $f' = 2f$. To retain the same magnific

Q: What are the options for this this exam Physics question?

A. increased two times · B. increased $\frac{3}{2}$ times · C. decreased two times · D. decreased $\frac{3}{2}$ times

Question

Q.42. A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be :-

FirstInTest · Accepted Answer

Sol. Magnification $M = \frac{L D}{F_o F_e}$. When the objective lens is cut vertically, its focal length changes. For a biconvex lens, $\frac{1}{f} = (n-1)(\frac{1}{R} + \frac{1}{R}) = (n-1)\frac{2}{R}$. When cut vertically, it becomes a plano-convex lens. For a plano-convex lens, $\frac{1}{f'} = (n-1)(\frac{1}{R} - \frac{1}{\infty}) = (n-1)\frac{1}{R}$. So, $f' = 2f$. To retain the same magnification, if $F_o$ becomes $2F_o$, then $L$ must also be doubled to keep $M$ constant. So, the tube length must be increased two times.

Answer

increased two times

Answer

increased $\frac{3}{2}$ times

Answer

decreased two times

Answer

decreased $\frac{3}{2}$ times

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