In an optics experiment, with the position of the object fixed, a student varies the position of a c

Question

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance $u$ and the image distance $v$, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of $45^{\circ}$ with the $x$-axis meets the experimental curve at $P$. The coordinates of $P$ will be\

Professor PC Thomas Classes · Answer

(a) : According to New Cartesian coordinate system used in our $12{ }^{	ext {th }}$ classes, for a convex lens, as $u$ is negative, the lens equation is\ $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$.\ One has to take that $u$ is negative again for calculation, it effectively comes to\ $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$\ If $u=$ radius of curvature, $2 f, v=2 f$ i.e., $\frac{1}{2 f}+\frac{1}{2 f}=\frac{1}{f}$.\ $v$ and $u$ are have the same value when the object is at the centre of curvature. The solution is (a).\ According the real and virtual system, $u$ is +ve and $v$ is also +ve as both are real. If $u=v$, $u=2 f=$ radius of curvature.\ $	herefore \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{2 f}+\frac{1}{2 f}=\frac{1}{f}$\ The answer is the same (a).\ (The figure given is according to New Cartesian system).