Question

The two slits in young's double slit experiment are separated by a distance of 0.03 mm when light of wavelength 5000 A 0 falls on the slits an interference pattern is produced on the screen 1.5 m away find the distance of the fourth bright fringe from the central maxima

Answer 1

fringe width = w = D λ /d = 1.5 m* 5000 * 10⁻¹⁰ m/3*10⁻⁵ m 
 w = 2.5 cm

distance of 4th bright fringe from central maxima = 4 * w or 4 Dλ/d
       = 10 cm.

Answer 2

Explanation:

Intensity is proportional to width of the slit. thus,

I

2

I

1

=

25

1

or

a

2

a

1

=

I

2

I

1

=

5

1

or a

2

=5a

1

now,

I

min

I

max

=

(a

1

−a

2

)

2

(a

1

+a

2

)

=

(a

1

−5a

2

)

2

(a

1

+5a

1

)

2

=36/16=9/4