A point object is placed at the centre of a g | vedclass

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

Question

(C) For refraction at a spherical surface,the formula is $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Here,the object is at the center,so $u

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) For refraction at a spherical surface,the formula is $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Here,the object is at the center,so $u = -R = -6 \, cm$.The light travels from glass $(n_1 = 1.5)$ to air $(n_2 = 1.0)$.The radius of curvature $R$ for the surface is $-6 \, cm$ (as the center of curvature is to the left of the surface).Substituting the values: $\frac{1.0}{v} - \frac{1.5}{-6} = \frac{1.0 - 1.5}{-6}$.$\frac{1}{v} + 0.25 = \frac{-0.5}{-6} = \frac{0.5}{6} = \frac{1}{12}$.$\frac{1}{v} = \frac{1}{12} - \frac{1}{4} = \frac{1 - 3}{12} = \frac{-2}{12} = -\frac{1}{6}$.Thus,$v = -6 \, cm$.The negative sign indicates that the image is virtual and formed on the same side as the object,at a distance of $6 \, cm$ from the surface.

$A$ point object is placed at the centre of a glass sphere of radius $6 \, cm$ and refractive index $1.5$. The distance of the virtual image from the surface of the sphere is.....$cm$.

Similar Questions

The area (in square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is

If $\sqrt{9 x^2+6 x+1} < (2-x)$,then:

$A$ ray of light strikes a plane mirror $M$ at an angle of $45^o$ as shown in the figure. After reflection,the ray passes through a prism of refractive index $1.5$ whose apex angle is $4^o$. The total angle through which the ray is deviated is....$^o$

What is an embryoid?

$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the

Vedclass Test Series

Exam Paper Generator

Online Exam Module