Question

A convex lens forms a real image. The image on the other side of the lens is three times farther from the lens than the object is from the lens. Now the object and screen are moved until the image is two times farther from the lens than the object. If the increase in distance of the object is 6 cm, then the shift of the screen is:

• A. 36 cm
• B. 72 cm
• C. 18 cm
• D. 9 cm

Answer 1

The correct option is A 36 cm

Let object and image distance be u and v respectively and focal length = f.
For the first case when image is three times farther from the lens than the object is from the lens
$3=\frac{v}{u}$
$\Rightarrow v=3u$
Using lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{3u}+\frac{1}{u}=\frac{1}{f}$
or $\frac{4}{3u}=\frac{1}{f}$ ….. (1)

For the case when object distance is increased by 6, the image beomes two times frather from the lens than the objec. So,
$2=\frac{v}{u+6}$
$\Rightarrow v=2(u+6)$
Using lens formula again,
$\frac{1}{2(6+u)}+\frac{1}{6+u}=\frac{1}{f}$
or $\frac{3}{2(6+u)}=\frac{1}{f}$ ….. (2)
Eqs. (1) and (2) gives
$\frac{4}{3u}=\frac{3}{2(6+u)}$
$\therefore 9u=48+8u$
$u=48cm$
Shift of screen = 3u – 2(6+u)
$=3\times 48–2(6+48)$
= 36 cm