Question

An object O is placed at a point distance $x_1$ from the focus $F_1$ of a converging lens of focal length f and its image I is formed at distance $x_2$ from the focus $F_2$ as shown in figure, then:

• A. magnitude of transverse magnification is $\left|\frac{f}{x_1}\right|$
• B. magnitude of transverse magnification is $\left|\frac{x_2}{f}\right|$
• C. $x_1\times x_2=f^2$
• D. $x_1+x_2\ge 2f$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

magnitude of transverse magnification is $\left|\frac{f}{x_1}\right|$

B

magnitude of transverse magnification is $\left|\frac{x_2}{f}\right|$

C

$x_1\times x_2=f^2$

D

$x_1+x_2\ge 2f$

Using lens formula, $u=–(f+x_1);v=f+x_2$

$\frac{1}{f+x_2}-\frac{1}{-(f+x_1)}=\frac{1}{f}$

$(2f+x_1+x_2)f=f^2+x_1{x}_2+f(x_1+x_2)f^2=x_1{x}_2$

$x_1+x_2=x_1+\frac{f^2}{x_1}$

$=\frac{x_1^2+f^2}{x_1}=\frac{(x_1-f{)}^2+2f{x}_1}{x_1}$

$x_1+x_2=\frac{(x_1-f{)}^2}{x_1}+2f\ge 2f$

$m=\frac{v}{u}=\frac{f+x_2}{-(f+x_1)}=\frac{\frac{f{x}_1+f^2}{x_1}}{-(f+x_1)}=\frac{-f}{x_1}$

Also, $m=\frac{f+x_2}{-(f+x_1)}=\frac{f+x_2}{-(f+\frac{f^2}{x_2})}=\frac{-x_2}{f}$