Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width
[tex]D\sin\theta=n\lambda[/tex]
[tex]D=\dfrac{n\lambda}{\sin\theta}[/tex]
Where, D = width of slit
[tex]\lambda[/tex] = wavelength
Put the value into the formula
[tex]D=\dfrac{n\lambda}{\sin\theta}[/tex]
Here, [tex]\sin\theta[/tex] should be maximum.
So. maximum value of [tex]\sin\theta[/tex] is 1
Put the value into the formula
[tex]D=\dfrac{1\times\lambda}{1}[/tex]
[tex]D=\lambda[/tex]
(b). If the minimum number is 50
Then, the width is
[tex]D=\dfrac{50\times\lambda}{1}[/tex]
[tex]D=50\lambda[/tex]
(c). If the minimum number is 1000
Then, the width is
[tex]D=\dfrac{1000\times\lambda}{1}[/tex]
[tex]D=1000\lambda[/tex]
Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
