1.

First diffraction minimum due to a single slit of width 1.0xx10^(-5) cm is at 30^@. Then wavelength of light used is,

Answer»

400 A
500 Å
600 A
700 Å

Solution :For DIFFRACTION MINIMA, d SIN `theta = nlambda`
`LAMBDA=(dsintheta)/(n)=(1xx10^(-5)xxsin30^(@))/(1)=0.5xx10^(-7)`
`lambda=500A`


Discussion

No Comment Found

Related InterviewSolutions