Question

An object $2.4 \, m$ in front of a lens forms a sharp image on a screen $12 \, cm$ behind the lens. A glass plate $1 \, cm$ thick, of refractive index $1.50$ is interposed between lens and screen with its plane faces parallel to screen. At what distance from the lens, should object be placed so as to get a sharp image on the screen?

• A. $2.4 \, m$
• B. $3.2m$
• C. $5.6 \, m$
• D. $7.2m$

Answer 1

Answer: (C)

According to the thin lens formula
$\frac{1}{ f } = \frac{1}{ v ⁡} - \frac{1}{ u ⁡}$
Here, $u \, = \, -2.4 \, m \, = \, - \, 240 \, cm, \, v \, = \, 12 \, cm$
$\therefore $ $\frac{1}{ f } = \frac{1}{1 2} - \frac{1}{\left(- 2 4 0\right)} = \frac{1}{1 2} + \frac{1}{2 4 0}$
$\frac{1}{f }=\frac{2 1}{2 4 0}\text{ or }f⁡=\frac{2 4 0}{2 1} \, \text{cm}$
When a glass plate is interposed between lens and film, so shift produced by it will be
Shift $=\text{t }\left(1 - \frac{1}{\mu }\right)=1\left(1 - \frac{1}{1 \text{.} 5}\right)=1\left(1 - \frac{2}{3}\right)=\frac{1}{3} \, \text{cm}$
To get an image at the film, the lens should form an image at a distance
$v ^{'}=12-\frac{1}{3}=\frac{3 5}{3} \, \text{cm}$
Again using lens formula
$\therefore $ $\frac{2 1}{2 4 0} = \frac{3}{3 5} - \frac{1}{ u ^{'}} \text{ or } \frac{1}{ u ⁡^{'}} = \frac{3}{3 5} - \frac{2 1}{2 4 0} = \frac{1}{5} \left[\frac{3}{7} - \frac{2 1}{4 8}\right]$
$\frac{1}{ u ^{'}} = \frac{1}{5} \left[\frac{1 4 4 - 1 4 7}{3 3 6}\right] \text{ or } \frac{1}{ u ⁡^{'}} = - \frac{3}{1 6 8 0}$
$u ^{'}=-560 \, \text{cm}=-5\text{.}6 \, \text{m}$
$\left|u ^{'}\right|=5\text{.}6 \, \text{m}$