Light from a point source in air falls on a spherical glass surface (refractive index, μ=1.5 and radius of curvature =50 ~cm ). The image is formed at a distance of 200 cm from the glass surface inside the glass. The

Correct answer: 0 aligned & μ_2 v - μ_1 u = μ_2-μ_1 R & (1.5)/(200)- 1 -x =(1.5-1)/(50) aligned aligned & (1)/(x)=(1)/(100)-(3)/(400) & x=400 ~cm & x=4 m aligned

JEE Main 20253 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Evening Shift
Subject: Physics
Chapter: Ray Optics

JEE Main 2025 · 3 April · Evening Shift · Physics · Ray Optics

Light from a point source in air falls on a spherical glass surface...

Light from a point source in air falls on a spherical glass surface (refractive index,

\mu=1.5

and radius of curvature

=50 \mathrm{~cm}

). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ____ m. $Light from a point source in air falls on a spherical glass surface (refractive index, μ=1.5 and radius of curvature$

Correct answer

Step-by-step solution

$Light from a point source in air falls on a spherical glass surface (refractive index, μ=1.5 and radius of curvature$

\begin{aligned} & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{200}-\frac{1}{-\mathrm{x}}=\frac{1.5-1}{50}\end{aligned}

\begin{aligned} & \frac{1}{x}=\frac{1}{100}-\frac{3}{400} \\ & x=400 \mathrm{~cm} \\ & x=4 m\end{aligned}

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