Question

In a biprism experiment, the distance between the slit and eye-piece is 1 m. If the convex lens is interposed at a distance of 30 cm from the slit, then the size of the magnified image is 0.7 mm. The distance between the two virtual images of the slits will be:

• A. 0.1 mm
• B. 0.2 mm
• C. 0.3 mm
• D. 0.4 mm

Answer 1

The correct option is C 0.3 mm

$\begin{array}{c}\text{Lens formula}\\ \begin{array}{cc}\frac{1}{f}=\frac{1}{v}-\frac{1}{u}& \\ \text{magnification}& =\frac{v}{u}=\frac{h_i}{h_0}\\ & =\frac{70}{-30}=\frac{h_i=\left(0.7\right)}{h_0}\\ \Rightarrow & h_0=0.3mm\end{array}\end{array}$

So correct Answer= (C)