In a single-slit diffraction experiment,the f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a single-slit diffraction pattern,the condition for minima is given by $d \sin 	heta = n \lambda$,where $n = 1, 2, 3, ...$.For the first mini

Vedclass · Accepted Answer

$440$

Vedclass · Answer

(A) For a single-slit diffraction pattern,the condition for minima is given by $d \sin 	heta = n \lambda$,where $n = 1, 2, 3, ...$.For the first minimum $(n=1)$,we have $d \sin 	heta_1 = \lambda_1$,which implies $\sin 	heta_1 = \frac{\lambda_1}{d}$.For the first secondary maximum,the condition is $d \sin 	heta_2 = (n + \frac{1}{2}) \lambda_2$. For the first secondary maximum,$n=1$,so $d \sin 	heta_2 = \frac{3}{2} \lambda_2$,which implies $\sin 	heta_2 = \frac{3 \lambda_2}{2d}$.Since the first minimum of $\lambda_1$ coincides with the first maximum of $\lambda_2$,we have $	heta_1 = 	heta_2$,hence $\sin 	heta_1 = \sin 	heta_2$.Equating the two expressions: $\frac{\lambda_1}{d} = \frac{3 \lambda_2}{2d}$.Solving for $\lambda_2$: $\lambda_2 = \frac{2}{3} \lambda_1$.Substituting $\lambda_1 = 660 \ nm$: $\lambda_2 = \frac{2}{3} 	imes 660 \ nm = 440 \ nm$.

In a single-slit diffraction experiment,the first minimum for a wavelength $\lambda_1 = 660 \ nm$ coincides with the first maximum for a wavelength $\lambda_2$. Find the value of $\lambda_2$ in $nm$.

Similar Questions

$A$ screen is placed at a distance $50\,cm$ from a single slit,which is illuminated with light of wavelength $690\,nm$. If the distance between the first and third minima is $3.00\,mm$,what is the width of the slit (in $,mm$)?

$A$ screen is placed $0.5 \,m$ away from a single slit which is illuminated by a monochromatic light of wavelength $6000 \text{ Å}$. If the distance between the first and third minima in the diffraction pattern on the screen is $3 \,mm$, then the slit width is: (in $\,mm$)

Red light is generally used to observe a diffraction pattern. If green light is used instead of red light,the diffraction pattern will.....

If a classroom door is open just a small amount, we can hear sounds coming from the room but we can't see what is going on inside the room because

Assertion : Diffraction takes place for all types of waves,mechanical or non-mechanical,transverse or longitudinal.
Reason : Diffraction effects are perceptible only if the wavelength of the wave is comparable to the dimensions of the diffracting device.

Vedclass Test Series

Exam Paper Generator

Online Exam Module