Question

In a single slit diffraction experiment, first minimum for a light of wavelength 480 nm coincides with the first maximum of another wavelength `lambda.` Then `lambda'` is ____________.

Options

• 3200 Å

• 2400 Å

• 4800 Å

• 5200 Å

Answer 1

In a single slit diffraction experiment, first minimum for a light of wavelength 480 nm coincides with the first maximum of another wavelength `lambda.` Then `lambda'` is 3200 Å.

Explanation:

`"For minima, d sin"  theta = "n" lambda`

Here n = 1,

`"d" ("y"_(1"d")/"D") = 1(4800  Å )`

`"y"_(1"d") = "D"/"d" (4800  Å)`

Now, first maximum is approximately between the first minima and second minima.

`"y"_(1"b") = (("y"_(1"d") + "y"_(2"d"))/2) = ((1 + 2)/2) ("D"lambda')/"d"`

`"As"  "y"_(1"d") = "y"_(1"b") Rightarrow "D"/"d" (4800  Å)`

` = (3/2)"D"/"d" lambda'`

`therefore lambda'  = (2 xx 4800 Å)/3`

` = 3200 Å`