Question

A Lloyd’s mirror of length 5 cm is illuminated with monochromatic light of wavelength $\lambda (=6000\stackrel{\circ }{A})$ from a narrow 1 mm slit in its plane and $5\text{cm}$ plane from its near edge. Find the fringe width on a screen $120\text{cm}$ from the slit and width of interference pattern on the screen.

• A. $0.036cm,1.2cm$
• B. $0.045cm,0.96cm$
• C. $1.2cm,0.036cm$
• D. $3.6cm,0.12cm$

Answer 1

The correct option is A $0.036cm,1.2cm$


Fringe width ($\beta$)=$\frac{\lambda D}{d}$
from the given information,
$\beta$ =$\frac{6000\times {10}^{-8}\times 120}{0.2}cm\beta =36\times {10}^{-3}cm$

Width of interference pattern on the screen:

$\Delta PAO$
$tan{\theta }_1=\frac{AO}{115}$ ...(1)

$\Delta MBO$
$tan{\theta }_2=\frac{BO}{110}$ ...(2)

From the origin side,
$tan{\theta }_1=\frac{0.1}{5}$
$tan{\theta }_2=\frac{0.1}{10}$
putting these values on eq.(1) and eq.(2) respectively.
so, $tan{\theta }_1=\frac{AO}{115}\frac{0.1}{5}=\frac{AO}{115}AO=\frac{0.1\times 115}{5}=2.3$
and $tan{\theta }_2=\frac{BO}{110}\frac{0.1}{10}=\frac{BO}{110}BO=\frac{0.1\times 110}{10}=1.1$

Now, width of interference pattern on the screen = $AO-BO$
$2.3-2.2=1.2cm$