A convex lens of diameter 5.0 cm and focal length $\mathrm{f}=20.0\mathrm{cm}$ was cut along the diameter into two identical halves. In the process, a thin layer of lens (a = 1.0 mm) is lost. Then the halves were put together to form a composite lens. A source slit is placed 10 cm from the lens, emitting light of wavelength $\lambda =600\mathrm{nm}$ Behind the lens, a screen wall located at a distance b = 80 cm from lens. Find the fringe width :

The source will act as an extended object to the lenses separately , with an object height of 0.5mm (half of the distance between the lenses)

The two lenses will form two images separately.

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{20}-\frac{1}{10}$

$\Rightarrow \mathrm{v}=-20$

So each image will be at a distance of 20cm from each lens, and the magnification is '2'.

The images formed are at a distance of 1mm from each other.

Images of the source is formed at 100 cm from the screen.

Fringe width is given by $\mathrm{\beta }=\frac{\mathrm{D\lambda }}{\mathrm{d}}=\frac{1\times 600\times {10}^{-9}}{{10}^{-3}}=0.6\mathrm{mm}$