Question

An object is kept at a distance of $16cm$ from a thin lens and the image formed is real. If the object is kept at a distance of $6cm$ from the same lens, the image formed is virtual. If the sizes, of the images formed are equal, the focal length of the lens will be

• A. $15cm$
• B. $17cm$
• C. $21cm$
• D. $11cm$

Answer 1

The correct option is D $11cm$

When object is kept 16 cm away,

$\frac{1}{v_1}-\frac{1}{(-16)}=\frac{1}{f}$............ (1)

Magnification=$\frac{v_1}{(-16)}$

When object is kept 6 cm away,

$\frac{1}{v_2}-\frac{1}{(-6)}=\frac{1}{f}$................ (2)

Magnification=$\frac{v_2}{-6}$

Since the sizes of the images is same, $M_1=-M_2$

Hence, $v_2=\frac{3v_1}{8}$

Using this relation in Equation 1 and Equation 2, we get

$f=11cm$