Question

An object $2.4\text{m}$ in front of a lens forms a sharp image on a film $12\text{cm}$ behind the lens. A glass plate $1\text{cm}$ thick, of refractive index $1.50$ is interposed between lens and film with its plane faces parallel to film. By what distance (from lens) in m should object be shifted to form a sharp image on film?

• A. 3
• B. 3.2
• C. 3.20

Answer 1

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

Shift in image position due to glass plate,
S = $(1-\frac{1}{\mu })t=(1-\frac{1}{1.5})\times 1=\frac{1}{3}\text{cm}$
For focal length of the lens,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{12}-\frac{1}{-240}$
or $\frac{1}{f}=\frac{20+1}{240}$
$\Rightarrow f=\frac{240}{21}cm$
Now, to get back image on the film, lens has to form image at
$(12-\frac{1}{3})\text{cm}$ = $\frac{35}{3}\text{cm}$ such that the glass plate will shift the image on the film.
As $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{3}{35}-\frac{21}{240}$
$=\frac{48\times 3-7\times 21}{1680}$
$=-\frac{1}{560}$
$\Rightarrow u=-5.6\text{m}$
Hence object should be shifted by $5.6-2.4=3.2\text{m}$

Ans: 5.6

Why this question? Caution: Since the slab shifts the image farther away from the lens, in order to form the image again on the film, the image distance should be less than the original image distance by the amount of shift caused.