Question 445: Wave Optics

Question

A light source, width emits two wavelengths $$\lambda$$
1
= 400 nm and $$\lambda$$
2
= 600 nm, is used in a Young's double-slit experiment. If recorded fringe widths for $$\lambda$$
1
and $$\lambda$$
2
are $$\beta$$
1
and $$\beta$$
2
and the number of fringes for them within a distance y on one side of the central maximum are m
1
and m
2
, respectively, then

Accepted Answer

Fringe width $$\beta  = {{\lambda D} \over d}$$; $$\lambda$$
1
= 400 nm; $$\lambda$$
2
= 400 nm
Since $$\lambda$$
2
> $$\lambda$$
1
, $$\beta$$
2
> $$\beta$$
1
Number of fringes within a distance y is given by
$${m_1} = {y \over {{\beta _1}}}$$
$${m_2} = {y \over {{\beta _2}}}$$
Since, $$\beta$$
2
> $$\beta$$
1
, we get m
2
 $$\beta$$
1
$$\Rightarrow$$ $$	heta$$
2
> $$	heta$$
1

Answer

$$\beta$$
2
> $$\beta$$
1

Answer

m
1
> m
2

Answer

from the central maximum, 3rd maximum of $$\lambda$$
2
overlaps with 5th minimum of $$\lambda$$
1

Answer

the angular separation of fringes of $$\lambda$$
1
is greater than $$\lambda$$
2

Wave Optics

Choose the correct answer: