On a hypotenuse of a right prism (30°-60°-90°) of refractive index 1.50, a drop of liquid is placed as shown in figure. Light is allowed to fall normally on the short face of the prism. In order that the ray of light may get totally reflected, the maximum value of refractive index is :

Question

On a hypotenuse of a right prism (30°-60°-90°) of refractive index 1.50, a drop of liquid is placed as shown in figure. Light is allowed to fall normally on the short face of the prism. In order that the ray of light may get totally reflected, the maximum value of refractive index is : (A) 1.30 (B) 1.47 (C) 1.20 (D) 1.25

Tardigrade Admin · Accepted Answer

for total internal infliction By snells low ( sin 60/ sin 90) = (M textliquid/1.50) (√3/2)× (3/2) = M textliquide M textliquide = 1.29 = 1.3

Q. On a hypotenuse of a right prism (30∘−60∘−90∘) of refractive index 1.50, a drop of liquid is placed as shown in figure. Light is allowed to fall normally on the short face of the prism. In order that the ray of light may get totally reflected, the maximum value of refractive index is :

1.30

1.47

1.20

1.25

for total internal infliction By snells low sin90sin60​=1.50Mliquid​​ 23​​×23​=Mliquide​ Mliquide​=1.29=1.3

for total internal infliction
By snells low
$\frac{s i n 60}{s i n 90} = \frac{M _{liquid}}{1.50}$
$\frac{3}{2} \times \frac{3}{2} = M_{liquide}$
$M_{liquide} = 1.29 = 1.3$