A point object O is placed in front of a glass rod having a spherical end of radius of curvature 30 cm- The image would be formed -30 cm leftInfinity1 cm to the right18 cm to the left

Question

Toppr Admin · Accepted Answer

-x3BC-1-air-1-x3BC-2-glass-1-5u-x2212-15cmR-30cm-x3BC-2-x2212-x3BC-1R-x3BC-2v-x2212-x3BC-1u1-5-x2212-130-1-5v-x2212-115v-x2212-30Ans- 30cm towards the left-

A point object $O$ is placed in front of a glass rod having a spherical end of radius of curvature $30 c m$ . The image would be formed at :

$30 c m$ left
Infinity
$1 c m$ to the right
$18 c m$ to the left

$\begin{matrix} μ_{1} (a i r) = 1 μ_{2} (g l a s s) = 1.5 u = - 15 c m R = 30 c m \frac{μ_{2} - μ_{1}}{R} = \frac{μ_{2}}{v} - \frac{μ_{1}}{u} \frac{1.5 - 1}{30} = \frac{1.5}{v} - \frac{1}{15} v = - 30 \end{matrix}$
Ans. $30 c m$ towards the left.

A point object O is placed in front of a glass rod having a spherical end of radius of curvature 30 cm. The image would be formed at :30 cm leftInfinity1 cm to the right18 cm to the left

μ1(air)=1μ2(glass)=1.5u=−15cmR=30cmμ2−μ1R=μ2v−μ1u1.5−130=1.5v−115v=−30Ans. 30cm towards the left.

A point object $O$ is placed in front of a glass rod having a spherical end of radius of curvature $30 c m$ . The image would be formed at :

$30 c m$ left
Infinity
$1 c m$ to the right
$18 c m$ to the left

$\begin{matrix} μ_{1} (a i r) = 1 μ_{2} (g l a s s) = 1.5 u = - 15 c m R = 30 c m \frac{μ_{2} - μ_{1}}{R} = \frac{μ_{2}}{v} - \frac{μ_{1}}{u} \frac{1.5 - 1}{30} = \frac{1.5}{v} - \frac{1}{15} v = - 30 \end{matrix}$
Ans. $30 c m$ towards the left.