Question 266: Wave Optics

Question

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If the distance between the first minimum on either side of the central maximum is 1 cm , the wavelength of light used is

Accepted Answer

Given:
Slit width, $a = 0.2\, 	ext{mm} = 2 	imes 10^{-4}\, 	ext{m}$
Distance from slit to screen, $D = 2\, 	ext{m}$
Distance between the first minima on both sides = $1\, 	ext{cm} = 0.01\, 	ext{m}$
Let the wavelength of light be $\lambda$.
Step 1: Condition for minima in single-slit Fraunhofer diffraction
For the first minimum (on either side), the angular position $	heta$ satisfies:
$$ a \sin	heta = m\lambda, \quad 	ext{where } m = \pm 1 $$
Step 2: Small angle approximation
For small $	heta$, $\sin	heta \approx 	an	heta \approx 	heta$.
The linear distance from the center to the first minimum on the screen,
$$ y_1 = D 	an	heta \approx D \sin	heta \approx D 	heta $$
So,
$$ \sin	heta = \frac{y_1}{D} $$
Step 3: Distance between first minima on both sides
First minimum appears at $y_1$ above and $-y_1$ below the central maximum, so the total distance is
$$ Y = 2y_1 $$
Given $Y = 0.01\, 	ext{m}$,
$$ y_1 = \frac{Y}{2} = \frac{0.01}{2} = 0.005\,	ext{m} $$
Step 4: Substitute and solve
Using $a \sin	heta = \lambda$,
$$ a \sin	heta = \lambda $$
$$ \lambda = a \sin	heta = a \frac{y_1}{D} $$
Substitute the values:
$a = 2 	imes 10^{-4}~	ext{m}$
$y_1 = 0.005~	ext{m}$
$D = 2~	ext{m}$
So,
$$ \lambda = 2 	imes 10^{-4} 	imes \frac{0.005}{2} $$
$$ \lambda = 2 	imes 10^{-4} 	imes 0.0025 $$
$$ \lambda = 5 	imes 10^{-7}~	ext{m} $$
Step 5: Convert to Angstrom
$1~	ext{ÃÂ} = 10^{-10}~	ext{m}$
$$ \lambda = 5 	imes 10^{-7}~	ext{m} = 5 	imes 10^{-7}~	ext{m} 	imes \frac{1~	ext{ÃÂ}}{10^{-10}~	ext{m}} = 5 	imes 10^{3}~	ext{ÃÂ} = 5000~	ext{ÃÂ} $$
Final Answer
Option
C
: $5000~\mathring{A}$

Answer

$2000 \mathop {\rm{A}}\limits^{\rm{^\circ }}$

Answer

$4000 \mathop {\rm{A}}\limits^ \circ$

Answer

$5000 \mathop {\rm{A}}\limits^ \circ$

Answer

$10000 \mathop {\rm{A}}\limits^ \circ$

Wave Optics

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If the distance between the first minimum on either side of the central maximum is 1 cm , the wavelength of light used is

Choose the correct answer: