From a point source a light falls on a spherical glass surface (μ = 1.5 and radius of curvature = 10 cm). The distance between point source and glass surface is 50 cm. The position of image is

Question

From a point source a light falls on a spherical glass surface (μ = 1.5 and radius of curvature = 10 cm). The distance between point source and glass surface is 50 cm. The position of image is (A) 25 cm (B) 50 cm (C) 100 cm (D) 150 cm

Tardigrade Admin · Accepted Answer

Using, (μ2/v)-(μ1/u)=( μ2-μ1/R) Here, μ1=1, μ2 = 1.5, u = -50 cm ∴ (1.5/v) - (1/(-50)) = ((1.5-1)/10) ⇒ (1.5/v) = 0.05 - 0.02 = 0.03 ⇒ v = 50 cm

Q. From a point source a light falls on a spherical glass surface $(μ = 1.5$ and radius of curvature $= 10 c m$ ). The distance between point source and glass surface is $50 c m$ . The position of image is

$25 c m$

$50 c m$

$100 c m$

$150 c m$

Using, $\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$
Here, $μ_{1} = 1, μ_{2} = 1.5, u = - 50 c m$
$∴ \frac{1.5}{v} - \frac{1}{( - 50 )}$
$= \frac{( 1.5 - 1 )}{10}$
$\Rightarrow \frac{1.5}{v}$
$= 0.05 - 0.02 = 0.03$
$\Rightarrow v = 50 c m$

Q. From a point source a light falls on a spherical glass surface (μ=1.5 and radius of curvature =10cm). The distance between point source and glass surface is 50cm. The position of image is

25cm

50cm

100cm

150cm