Light from a point source in air falls on a spherical glass surface (refractive index, = 1.5 and radius of curvature = 50 cm). The image is formed at a distance of 200 cm from the light source from the glass surface is ________ m. [2025]

Question

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 magnitude of distance of the light source from the glass surface is ________ m. [2025]

Smart Achievers · Accepted Answer

(4)

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.5}{200} - \frac{1}{- x} = \frac{1.5 - 1}{50}$

$\frac{1}{x} = \frac{1}{100} - \frac{3}{400}$

$\Rightarrow x = 400 cm$ $= 4 m$

Solution

Ans.
(4)

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.5}{200} - \frac{1}{- x} = \frac{1.5 - 1}{50}$

$\frac{1}{x} = \frac{1}{100} - \frac{3}{400}$

$\Rightarrow x = 400 cm$ $= 4 m$

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Solution

Ans. (4) μ2v–μ1u=μ2–μ1R 1.5200–1–x=1.5–150 1x=1100–3400 ⇒x=400 cm=4 m

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Ans.
(4)

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.5}{200} - \frac{1}{- x} = \frac{1.5 - 1}{50}$

$\frac{1}{x} = \frac{1}{100} - \frac{3}{400}$

$\Rightarrow x = 400 cm$ $= 4 m$