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Question

In a Youngs double experiment, the slits are $$1.5\ mm$$ apart. When the slits are illuminated by a monochromatic light source and the screen is kept $$1\ m$$ apart from the slits, width of $$10$$ fringes is measured as $$3.93\ mm$$. Calculate the wavelength of light used. What will be the width of $$10$$ fringes when the distance between the slits and the screen is increased by $$0.5\ m$$. The source of light used remains the same. 


Solution

We know,

Fringe width=$$\dfrac{\lambda D}{d}$$

Here fringe width in 1st case =$$\dfrac{3.93}{10}=0.393 mm$$

$$D=1000mm$$, $$d=1.5mm$$

$$\lambda=\dfrac{0.393\times 1.5}{1000}=\dfrac{3.393\times 1.5}{10000}$$

$$\lambda=5.0895\times 10^{-4}mm$$

$$=508.95 nm$$

Now , fringe width of 10 fringes when $$d=1.5+0.5=2.0m$$

Let initial and final fringe width of 10 fringes be $$f_i,f_f$$

$$\dfrac{f_i}{f_f}=\dfrac{d_f}{d_i}=\dfrac{2}{1.5}=\dfrac{4}{3}$$


$$f_f=\dfrac{3.93\times 4}{3}=52.4mm$$


Physics

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