A denser medium of refractive index 1.5 has a | vedclass

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

Question

(B) The formula for refraction at a spherical surface is given by: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,the light travel

Vedclass · Accepted Answer

$A$ virtual image at $4.8 \, cm$

Vedclass · Answer

(B) The formula for refraction at a spherical surface is given by: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,the light travels from the denser medium $(\mu_1 = 1.5)$ to air $(\mu_2 = 1.0)$.The object distance $u = -9 \, cm$ (measured against the direction of incident light).The radius of curvature $R = +12 \, cm$ (since the center of curvature is on the side of the incident light for a concave surface viewed from the denser medium).Substituting the values:$\frac{1.0}{v} - \frac{1.5}{-9} = \frac{1.0 - 1.5}{12}$$\frac{1}{v} + \frac{1.5}{9} = \frac{-0.5}{12}$$\frac{1}{v} + \frac{1}{6} = -\frac{1}{24}$$\frac{1}{v} = -\frac{1}{24} - \frac{1}{6} = \frac{-1 - 4}{24} = -\frac{5}{24}$$v = -\frac{24}{5} = -4.8 \, cm$.The negative sign indicates that the image is formed on the same side as the object,making it a virtual image at a distance of $4.8 \, cm$ from the pole. Thus,option $(b)$ is correct.

$A$ denser medium of refractive index $1.5$ has a concave surface of radius of curvature $12 \, cm$. An object is situated in the denser medium at a distance of $9 \, cm$ from the pole. Locate the image due to refraction in air.

Similar Questions

As shown in the figure,region $BCDEF$ and $ABFG$ have refractive indices $2.0$ and $1.5$ respectively. $A$ particle $O$ is kept at the midpoint of region $BCEF$. The final image of the object as seen by the eye is at a distance of:

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:

Derive the relation between object distance $(u)$,image distance $(v)$,refractive indices of the media ($n_1$ and $n_2$),and the radius of curvature $(R)$ for a spherical refracting surface.

$A$ curved surface of radius $R$ separates two media of refractive indices $\mu_1$ and $\mu_2$ as shown in figures $A$ and $B$. Choose the correct statement$(s)$ related to the real image formed by the object $O$ placed at a distance $x$,as shown in figure $A$.

$A$ parallel beam of light travelling in water (refractive index $\mu_1 = 4/3$) is refracted by a spherical air bubble of radius $R = 2 \, cm$ situated in water. Assuming the light rays to be paraxial,the position of the image due to refraction at the first surface is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module