A spherical glass is attached to a rigid wall | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The light rays from point $A$ on the wall travel through the glass and refract at the spherical surface into the air.For refraction at a spherical

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(D) The light rays from point $A$ on the wall travel through the glass and refract at the spherical surface into the air.For refraction at a spherical surface, the formula is $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here, the light travels from glass $(\mu_1 = 1.5)$ to air $(\mu_2 = 1.0)$.The object distance $u$ is the distance of $A$ from the pole $X$. Since $XA = 3 	ext{ cm}$ and light travels from $A$ to $X$, $u = -3 	ext{ cm}$.The radius of curvature $R$ for the convex surface (as seen from the glass side) is $-5 	ext{ cm}$ because the center of curvature lies to the left of the pole $X$.Substituting these values into the formula:$\frac{1.0}{v} - \frac{1.5}{-3} = \frac{1.0 - 1.5}{-5}$$\frac{1}{v} + 0.5 = \frac{-0.5}{-5} = 0.1$$\frac{1}{v} = 0.1 - 0.5 = -0.4$$v = -\frac{1}{0.4} = -2.5 	ext{ cm}$.The negative sign indicates that the image $I$ is formed $2.5 	ext{ cm}$ to the left of the pole $X$.The distance of the observer $O$ from the pole $X$ is $OX = OA - XA = 8 - 3 = 5 	ext{ cm}$.The apparent distance of $A$ from the observer $O$ is $OI = OX + XI = 5 	ext{ cm} + 2.5 	ext{ cm} = 7.5 	ext{ cm}$.

Similar Questions

In a medium of refractive index $1.6$ having a convex surface,a point object is placed at a distance of $12 \,cm$ from the pole. The radius of curvature of the surface is $6 \,cm$. Locate the image as seen from air.

$A$ parallel beam of light is allowed to fall on a transparent spherical globe of diameter $30 \,cm$ and refractive index $1.5$. The distance from the centre of the globe at which the beam of light can converge is . . . . . . $mm$.

The figure shows a transparent sphere of radius $R$ and refractive index $\mu$. An object $O$ is placed at a distance $x$ from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface. If $x = 2R$,then the value of $\mu$ is

Write the equation for the image formed by a curved surface of radius of curvature $R$ in a medium of refractive index $n_1$.

Parallel rays are incident on a transparent sphere along its one diameter. After refraction,these rays converge at the other end of this diameter. The refractive index of the sphere is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module