Question

Consider red light with a wavelength of 690 nm in air that is incident onto two slits such that it gives diffraction fringes that are 1.00 mm apart on a screen at a distance L from the slits. If the screen is moved back by an additional 8.00 cm, the fringes become 1.40 mm apart. What is the separation between the slits? (a) 0.109 mm, (b) 0.138 mm, (c) 0.177 mm, (d) 0.218 mm, (e) 0.276 mm.

Answer 1

Option (b) 0.138 mm is the right answer.

Explanation:

Let  ' d ' be the distance between the two slits. Let the wavelength be ' λ '. Let, a screen be placed parallel to slits at a distance ' D ' from the slits. Let, ' β ' be the Bandwidth (also called Fringe width. ie Seperation between fringes.)

Consider, D₁ & D₂ as distance between screen and slits for the first case and the second case respectively. Similarly, take β₁ &   β₂ as the corresponding bandwidth (ie, seperation between the fringes) for the first case and second case respectively. Here, d & λ remains same.

Then, Given,

λ = 690 × 10^-9 m

D = L meter &   β₁ = 1 × 10^-3 meter -------------------------------- 1^st Case

D = L + (8 × 10^-2) meter &   β₂ = 1.4 × 10^-3 meter --------------------------------- 2^nd Case

d = ? , (The seperation between slits )

We have the equation of the bandwidth (β) as follows:

β = ( D/d ) × λ ------------------------------- ( Eqn 1 )

Substituting the 1^st Case values in the above Eqn 1,

   10^-3 = ( L / d ) × 690 × 10^-9      -------------------------------- ( Eqn 2 )

Similarly, Substituting the 2^ndCase values in the Eqn 1,

1.4 × 10^-3 = { [ L + (8 × 10^-2) ] / d } × 690 × 10^-9    -------------------------------- ( Eqn 3 )

Dividing ( Eqn 3 ) / ( Eqn 2 ),

1.4 =   ( L + (8 × 10^-2) ) / L

ie, 1.4 × L =   L + (8 × 10^-2)

ie,    1.4 L =   L + (8 × 10^-2)

ie, 1.4 L - L =   8 × 10^-2

ie,    0.4 L =   8 × 10^-2

ie, L = (8 × 10^-2) / 0.4

ie,    L = 0.2   

Now, Put the value of L (above) to ( Eqn 1 ) with any one of the case ( Case 1 OR case 2) to obtain ' d ',

For simplicity take the first case,

10^-3 = ( 0.2 / d ) × 690 × 10^-9

ie, d = ( 0.2 / 10^-3 ) × 690 × 10^-9

ie, d = 138 × 10^-6  meter

ie, d = 138 × 10^-3 mm

ie,   d = 0.138 mm

Therefore, option (b) 0.138 mm is the right answer.

THE SEPERATION BETWEEN THE SLITS IS d = 0.138 mm .