A thin film of soap solution (n = 1.4) lies on the top of a glass plate (n = 1.5). When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 and 630 nm. The minimum thickness of the soap solution is

Question

A thin film of soap solution (n = 1.4) lies on the top of a glass plate (n = 1.5). When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 and 630 nm. The minimum thickness of the soap solution is (A) 420 nm (B) 450 nm (C) 630 nm (D) 1260 nm

Tardigrade Admin · Accepted Answer

n1 λ1=n2 λ2 ∴ n1 × 420 =n2 × 630 or 2 n1=3 n2 If n2=2 then n1=3 Therefore, thickness of soap solution is given by or μ1 t =n1 (λ1/2) t =(3 × 420/1.4 × 2)=450 nm

Q. A thin film of soap solution $(n = 1.4)$ lies on the top of a glass plate $(n = 1.5)$ . When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths $420$ and $630 nm$ . The minimum thickness of the soap solution is

420 nm

450 nm

630 nm

1260 nm

$n_{1} λ_{1} = n_{2} λ_{2}$
$∴ n_{1} \times 420 = n_{2} \times 630$
or $2 n_{1} = 3 n_{2}$
If $n_{2} = 2$ then $n_{1} = 3$
Therefore, thickness of soap solution is given by
or $μ_{1} t = n_{1} \frac{λ _{1}}{2}$
$t = \frac{3 \times 420}{1.4 \times 2} = 450 nm$

Q. A thin film of soap solution (n=1.4) lies on the top of a glass plate (n=1.5). When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 and 630nm. The minimum thickness of the soap solution is