A student plots a graph between the inverse of magnification (1/m) produced by a thin convex lens and the object position u as shown in the figure. The focal length of the lens used, was

Question

A student plots a graph between the inverse of magnification (1/m) produced by a thin convex lens and the object position u as shown in the figure. The focal length of the lens used, was <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-0sx94l0jyqfb.png" /> (A) (b/ca) (B) (bc/a) (C) (c/b) (D) (b/c)

Tardigrade Admin · Accepted Answer

(1/v)-(1/u)=(1/f) (u/v)-1=(u/f) (u/v)=(u/f)+1 (1/m)=(u/f)+1 slope=(1/f) (1/f)=(b/c) f=(c/b)

Q. A student plots a graph between the inverse of magnification $\frac{1}{m}$ produced by a thin convex lens and the object position $u$ as shown in the figure. The focal length of the lens used, was

$\frac{b}{c a}$

$\frac{b c}{a}$

$\frac{c}{b}$

$\frac{b}{c}$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{u}{v} - 1 = \frac{u}{f}$
$\frac{u}{v} = \frac{u}{f} + 1$
$\frac{1}{m} = \frac{u}{f} + 1$
$s l o p e = \frac{1}{f}$
$\frac{1}{f} = \frac{b}{c}$
$f = \frac{c}{b}$

Q. A student plots a graph between the inverse of magnification m1​ produced by a thin convex lens and the object position u as shown in the figure. The focal length of the lens used, was

cab​

abc​

bc​

cb​