An object is placed in front of a sphere of radius R at a distance x from the first surface. The value of x for which the light after refraction from the first surface becomes parallel to the axis is

Question

An object is placed in front of a sphere of radius R at a distance x from the first surface. The value of x for which the light after refraction from the first surface becomes parallel to the axis is (A) (R/(μ-1)) (B) (R/μ+1) (C) (2 R/(1+μ)) (D) (R μ/(μ+1))

Tardigrade Admin · Accepted Answer

Applying, Lens' maker formula, we get (μ2/v)-(μ1/u)=(μ2-μ1/R) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/763e3ae8499b55b493de03d01cb8b2e5-.png" /> ⇒ (μ/∞)-(1/(-x))=(μ-1/+R)=+(1/x)=(μ-1/R) ∴ x=+((R/μ-1))=(R/(μ-1))

Q. An object is placed in front of a sphere of radius $R$ at a distance $x$ from the first surface. The value of $x$ for which the light after refraction from the first surface becomes parallel to the axis is

$\frac{R}{( μ - 1 )}$

$\frac{R}{μ + 1}$

$\frac{2 R}{( 1 + μ )}$

$\frac{R μ}{( μ + 1 )}$

Applying, Lens' maker formula, we get
$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$

$\Rightarrow \frac{μ}{\infty} - \frac{1}{( - x )} = \frac{μ - 1}{+ R} = + \frac{1}{x} = \frac{μ - 1}{R}$
$∴ x = + (\frac{R}{μ - 1}) = \frac{R}{( μ - 1 )}$

Q. An object is placed in front of a sphere of radius R at a distance x from the first surface. The value of x for which the light after refraction from the first surface becomes parallel to the axis is

(μ−1)R​

μ+1R​

(1+μ)2R​

(μ+1)Rμ​