In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(= 1.5), the central fringe shifts by 0.2 cm. The value of t i $s ____$ cm.
Options

Question

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(= 1.5), the central fringe shifts by 0.2 cm. The value of t i $s ____$ cm.

Options

G1 Physics · Accepted Answer

A transparent sheet introduces additional optical path (n-1)t. For a central-fringe shift x, d × x ─── D = (n-1)t. Therefore, t = xd ──────── D(n-1) . Substitute x = 0.2 cm, d = 0.1 cm, D = 50 cm, n = 1.5. Thus, t = (0.2)(0.1) ─────────── 50(1.5-1) = 8 × 10⁻⁴ cm. Correct option: A.

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(= 1.5), the central fringe shifts by 0.2 cm. The value of t i s ____ cm.