Two parallel beams of light P and Q are incident normally on a prism and the transmitted rays are brought to focus with the help of a convergent lens as shown in Fig. 2.23. If the intensities of the upper and lower beams immediately after transmission from face AC are 4I abd I, respectively, find the resultant intensity at the focus.

Explanation

In case of interference,
`I = I_(1) + I_(2) (sqrt (I_(1) I_(2))) cos phi`
Here, `I_(1) = I and I_(2) = 4I.` So,
`I_(R) = 5I + 4I cos phi`
`= I [1 + 8 cos^(2) ((phi)/(2))]`
with ` phi = (2 pi)/(lambda) (Delta x)`
where `Delta x = [DG - EF ("in medium")]`
Now, from Fig. 2.23,
`FE ("in medium") = mu (FE) = mu d tan theta`, [as `FE = d tan theta`] (iv)
`DG = DE cos (90^(@) - r) = DE sin r`
or `DG = (d sin r/cos theta) ["as" DE = d//cos theta]`
or ` DG = mu tan theta ["as at" D, mu sin theta = 1 sin r]` (v)
Substituting the value of FE and DG from Eqs. (iv) and (v) in Eq. (iii), we get
`Delta x = mu d tan theta - mu d tan theta = 0`
So, Eq. (ii) gives ` phi = 0^@` Hence, from Eq. (i),
`I_(R) = I [1 + 8 cos^(2) 0^(@) ]= 9I`.