The image formed by the converging beam after refraction through the curved surface is formed at

Question

The image formed by the converging beam after refraction through the curved surface is formed at <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-al2mip3abybh.png" /> (A) x=40 cm (B)  textx = (4 0/3) textcm (C)  textx = - (4 0/3) textcm (D)  textx = (1 8 0/7) textcm

Tardigrade Admin · Accepted Answer

We know that; ( textÎ¼2/ textv) - ( textÎ¼1/ textu) = ( textÎ¼2 - textÎ¼1/ textR) (1/ textv) - (3/2 × 3 0) = (1 - (3/2)/+ 2 0) (1/ textv) = - (1/4 0) + (1/2 0) = + (1/4 0) v=40 cm .

Q. The image formed by the converging beam after refraction through the curved surface is formed at

$x = 40 c m$

$x = \frac{40}{3} cm$

$x = - \frac{40}{3} cm$

$x = \frac{180}{7} cm$

We know that;
$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$
$\frac{1}{v} - \frac{3}{2 \times 30} = \frac{1 - \frac{3}{2}}{+ 20}$
$\frac{1}{v} = - \frac{1}{40} + \frac{1}{20} = + \frac{1}{40}$
$v = 40 c m$ .

Q. The image formed by the converging beam after refraction through the curved surface is formed at

x=40cm

x=340​cm

x=−340​cm

x=7180​cm

We know that; vμ2​​−uμ1​​=Rμ2​−μ1​​ v1​−2×303​=+201−23​​ v1​=−401​+201​=+401​ v=40cm .

$x = 40 c m$

$x = \frac{40}{3} cm$

$x = - \frac{40}{3} cm$

$x = \frac{180}{7} cm$

We know that;
$\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$
$\frac{1}{v} - \frac{3}{2 \times 30} = \frac{1 - \frac{3}{2}}{+ 20}$
$\frac{1}{v} = - \frac{1}{40} + \frac{1}{20} = + \frac{1}{40}$
$v = 40 c m$ .