A convex lens made up of a material of refractive index μ 1 is immersed in a medium of refractive index μ 2 as shown in the figure. The relation between μ 1 and μ 2 is :

Question

A convex lens made up of a material of refractive index μ 1 is immersed in a medium of refractive index μ 2 as shown in the figure. The relation between μ 1 and μ 2 is : (A)  μ 1<μ 2  (B)  μ 1>μ 2  (C)  μ 1=μ 2  (D)  μ 1=√μ 2

Tardigrade Admin · Accepted Answer

In a medium of higher refractive index, a convex lens diverges a parallel beam of light, thus behaving a diverging lens. Hence, μ 1<μ 2.

Q. A convex lens made up of a material of refractive index $μ_{1}$ is immersed in a medium of refractive index $μ_{2}$ as shown in the figure. The relation between $μ_{1}$ and $μ_{2}$ is :

$μ_{1} < μ_{2}$

$μ_{1} > μ_{2}$

$μ_{1} = μ_{2}$

$μ_{1} = μ_{2}$

In a medium of higher refractive index, a convex lens diverges a parallel beam of light, thus behaving a diverging lens. Hence, $μ_{1} < μ_{2} .$

Q. A convex lens made up of a material of refractive index μ1​ is immersed in a medium of refractive index μ2​ as shown in the figure. The relation between μ1​ and μ2​ is :

μ1​<μ2​

μ1​>μ2​

μ1​=μ2​

μ1​=μ2​​

In a medium of higher refractive index, a convex lens diverges a parallel beam of light, thus behaving a diverging lens. Hence, μ1​<μ2​.

Q. A convex lens made up of a material of refractive index $μ_{1}$ is immersed in a medium of refractive index $μ_{2}$ as shown in the figure. The relation between $μ_{1}$ and $μ_{2}$ is :

$μ_{1} < μ_{2}$

$μ_{1} > μ_{2}$

$μ_{1} = μ_{2}$

$μ_{1} = μ_{2}$

In a medium of higher refractive index, a convex lens diverges a parallel beam of light, thus behaving a diverging lens. Hence, $μ_{1} < μ_{2} .$