Question

Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 490 nm occurs at an angle of θ = 90°. Thus, it is on the verge of being eliminated from the pattern because θ cannot exceed 90° in Eq. 35-14. (a) What least wavelength in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all of the visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change in separation is needed?

Answer 1

Answer:

(a) d = 1960nm

(b) The slit should be decreased.

(c) Δd = 360nm.

Explanation:

The double-slit interference is given by the following equation:

$d sin(\theta) = m \lambda$      (1)

where d: is the distance between slits, Θ: is the angle between the path of the light and the screen, m: is the order of the interference and λ: is the wavelength of the light.  

(a) To determine the least wavelength in the visible range in the third-order we need first to find the distance between slits, using equation (1) for a fourth-order:  

$d = \frac{m \lambda}{sin(\theta)} = \frac{4 \cdot 490nm}{sin(90)} = 1960nm$  

Now, we can find the least wavelength in the visible range in the third-order:

$\lambda = \frac{d sin(\theta)}{m} = \frac{1960nm sin(90)}{3} = 653nm$

So, the least wavelength in the visible range (400nm - 700nm) in the third-order is 653nm.    

(b) To eliminate all of the visible light in the fourth-order maximum means that the wavelength must be smaller than 400nm, and hence the slit separation should be decreased since they are proportional to each other (see equation (1)).    

(c) The distance between slits needed to eliminate all of the visible light in the fourth-order maximum, with λ = 400 nm as limit value, is:

$d = \frac{m \lambda}{sin(\theta)} = \frac{4 \cdot 400nm}{sin(90)} = 1600nm$  

Therefore the least change in separation needed is equal to the initial distance calculated for 490nm and the final distance calculated for 400nm:  

$\Delta d = d_{i} - d_{f} = 1960nm - 1600nm = 360nm$

I hope it helps you!