A slab of a glass of refractive index 1.5 and thickness 3 cm is placed with the faces perpendicular to the principal axes of a concave mirror. If the radius of curvature of the mirror is 10 cm , the distance at which an object must be placed from the mirror so that image coincides with the object is

Question

A slab of a glass of refractive index 1.5 and thickness 3 cm is placed with the faces perpendicular to the principal axes of a concave mirror. If the radius of curvature of the mirror is 10 cm , the distance at which an object must be placed from the mirror so that image coincides with the object is (A) 9cm (B) 10cm (C) 11 cm (D) 12cm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-kkoysaje9h4d.png" /> μ = text1.5 = 3 / 2 Δ textx = textt (1 - (1/μ )) = 3 [1 - (1/3 / 2)] = 3 ((1/3)) Δ textx = text1 cm object distance = 10 + 1 = 11 cm

$9 c m$

$10 c m$

$11 c m$

$12 c m$

$μ = 1.5 = 3/2$
$Δ x = t (1 - \frac{1}{μ}) = 3 [1 - \frac{1}{3/2}] = 3 (\frac{1}{3})$
$Δ x = 1 cm$
object distance = 10 + 1 = 11 cm

9cm

10cm

11cm

12cm

μ=1.5=3/2 Δx=t(1−μ1​)=3[1−3/21​]=3(31​) Δx=1 cm object distance = 10 + 1 = 11 cm

$9 c m$

$10 c m$

$11 c m$

$12 c m$

$μ = 1.5 = 3/2$
$Δ x = t (1 - \frac{1}{μ}) = 3 [1 - \frac{1}{3/2}] = 3 (\frac{1}{3})$
$Δ x = 1 cm$
object distance = 10 + 1 = 11 cm