The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at $\frac{2^{\text{rd}}}{3}$ of the distance of the object from the surface. The wavelength of light inside the surface is $\frac{2}{3}$ times the wavelength in air. The radius of the curved surface is $\frac{x}{13} m$. The value of '$x$' is ______

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Geometrical Optics

The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at

\frac{2 ^{rd}}{3}

of the distance of the object from the surface. The wavelength of light inside the surface is

\frac{2}{3}

times the wavelength in air. The radius of the curved surface is

\frac{x}{13} m

. The value of '

x

' is ______

Answer: 30

Frequency is unchanged on refraction, so

λ \propto \frac{1}{n}

; a wavelength inside the medium equal to

\frac{2}{3}

of that in air means

n_{2} = \frac{3}{2}

, with

n_{1} = 1

for air. The real image lies 10 m behind the surface, so

v = + 10

m, and since this is

\frac{2}{3}

of the object distance, the object stands 15 m in front of the surface, giving

u = - 15

m. For refraction at a single spherical surface,

\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

\frac{3}{20} + \frac{1}{15} = \frac{1/2}{R} \Rightarrow \frac{13}{60} = \frac{1}{2 R}

R = \frac{30}{13}

m and

x = 30

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer