Question

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

Answer 1

Step 1:Given:

object distance, $u=-2.4m=-240cm$

image distance, $v=12cm$

Step 2:Finding the reciprocal of the focal length of the lens:

According to the lens formula:

$$\begin{gathered} \begin{array}{l}\frac{1}{f}=\frac{1}{v}–\frac{1}{u}\end{array} \\ bysubstitutingthegivenvalues,weget \\ \begin{array}{l}\frac{1}{f}=\frac{1}{12}–\frac{1}{-240}\end{array} \\ \begin{array}{l}\Rightarrow \frac{1}{f}=\frac{1}{12}+\frac{1}{240}\end{array} \\ \begin{array}{l}\Rightarrow \frac{1}{f}=\frac{21}{240}cm\end{array} \end{gathered}$$

Step 3:Finding the shift produced by the glass plate:

When a glass plate is interposed between lens and film, so shift produced by it.

The shift $t$ is given by

$$\begin{gathered} t=\begin{array}{l}(1–\frac{1}{\mu })\end{array}where\mu istherefractiveindexoftheglassplate \\ bysubstitutingthegivenvalue,weget \\ \begin{array}{l}(1–\frac{1}{1.5})\end{array} \\ =\begin{array}{l}(1–\frac{2}{3})\end{array} \\ =\begin{array}{l}\frac{1}{3}cm\end{array} \end{gathered}$$

In order to get the image on the film, the lens should form an image at a distance:

$\begin{array}{l}v\text{'}=12–\frac{1}{3}=\begin{array}{l}\frac{35}{3}cm\end{array}\end{array}$

Step 4:Finding the required position of the object using the lens formula:

$$\begin{gathered} \begin{array}{l}\frac{1}{f}=\frac{1}{v\text{'}}–\frac{1}{u\text{'}}\end{array} \\ \begin{array}{l}\Rightarrow \frac{21}{240}cm=\frac{3}{35}cm–\frac{1}{u^{\text{'}}}\end{array} \\ \begin{array}{l}\Rightarrow \frac{1}{u\text{'}}=\frac{3}{35}cm–\frac{21}{240}cm\end{array} \\ \begin{array}{l}\Rightarrow \frac{1}{u\text{'}}=\frac{1}{5}[\frac{3}{7}cm–\frac{21}{48}cm]\\ \begin{array}{l}\Rightarrow \frac{1}{u^{\text{'}}}=\frac{1}{5}\left[\frac{144–147}{336}\right]\end{array}\\ \begin{array}{l}\Rightarrow u^{\text{'}}=-560cm=-5.6m\end{array}\end{array} \end{gathered}$$

Hence, the object should be at 5.6m in front of the lens.