Question

A parallel beam of light of wavelength 5890 Å falls normally on a slit of width 0.2 mm. Find the distance between the first minima on the two sides of the central maximum of the diffraction pattern observed on a screen placed in the focal plane of a convex lens of focal length 50 cm. The lens is placed quite close to the slit.

Answer 1

We know  `sin theta = lambda/"e"`

or `theta = lambda/"e"` radian

If x be the linear distance of the first minimum from the central maximum on the screen and f be the focal length of the lens.

Then, `theta = (x/"f")`radian

`x = ("f"lambda)/"e"`

The linear distance between the two minima on the two sides of the central maximum is:

`therefore 2x = (2"f"lambda)/"e"`

`= (2 xx 0.5 xx (5890 xx 10^-10))/(0.2 xx 10^-3)`

`= 2.945 xx 10^-3 "m"`

= 2.945 mm