Two concave refracting surfaces of equal radi | vedclass

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

Question

(B) The formula for refraction at a single spherical surface is $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.For surface $B$ (concave

Vedclass · Accepted Answer

$0.114$

Vedclass · Answer

(B) The formula for refraction at a single spherical surface is $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.For surface $B$ (concave,radius $-R$):Object distance $u = -R/2$,$\mu_1 = 1$ (air),$\mu_2 = 1.5$ (glass).$\frac{1.5}{v_B} - \frac{1}{-R/2} = \frac{1.5 - 1}{-R} \Rightarrow \frac{1.5}{v_B} + \frac{2}{R} = -\frac{0.5}{R}$.$\frac{1.5}{v_B} = -\frac{0.5}{R} - \frac{2}{R} = -\frac{2.5}{R} \Rightarrow v_B = -\frac{1.5 R}{2.5} = -0.6 R$.For surface $A$ (concave,radius $+R$):Object distance $u = -(R + R/2) = -1.5 R$,$\mu_1 = 1$,$\mu_2 = 1.5$.$\frac{1.5}{v_A} - \frac{1}{-1.5 R} = \frac{1.5 - 1}{R} \Rightarrow \frac{1.5}{v_A} + \frac{1}{1.5 R} = \frac{0.5}{R}$.$\frac{1.5}{v_A} = \frac{0.5}{R} - \frac{1}{1.5 R} = \frac{0.75 - 1}{1.5 R} = -\frac{0.25}{1.5 R} = -\frac{1}{6 R}$.$v_A = -1.5 R 	imes 6 = -9 R$.Wait,re-evaluating surface $A$: The object is at distance $1.5 R$ from $A$. The surface is concave towards the object,so $R$ is negative. $\frac{1.5}{v_A} - \frac{1}{-1.5 R} = \frac{1.5 - 1}{-R} \Rightarrow \frac{1.5}{v_A} + \frac{2}{3 R} = -\frac{0.5}{R}$.$\frac{1.5}{v_A} = -\frac{0.5}{R} - \frac{0.666}{R} = -\frac{1}{2 R} - \frac{2}{3 R} = -\frac{7}{6 R}$.$v_A = -\frac{1.5 	imes 6 R}{7} = -\frac{9}{7} R \approx -1.2857 R$.The images are formed at $0.6 R$ to the left of $B$ and $1.2857 R$ to the left of $A$. The distance between $A$ and $B$ is $2 R$. The distance of image $I_B$ from $A$ is $2 R - 0.6 R = 1.4 R$. The distance of image $I_A$ from $A$ is $1.2857 R$. The separation is $1.4 R - 1.2857 R = 0.1143 R$.

Two concave refracting surfaces of equal radii of curvature $R$ and refractive index $1.5$ face each other in air as shown in the figure. $A$ point object $O$ is placed midway between $P$ and $B$. The separation between the images of $O$ formed by each refracting surface is: (in $R$)

Similar Questions

In the figure shown,$O$ is the centre of the glass sphere. The spot $P$ on the sphere,when viewed almost normally,appears:

$A$ spherical surface separates two media of refractive indices $1$ and $1.5$ as shown in the figure. Find the distance of the image of an object $O$ from the spherical surface. ($C$ is the center of curvature of the spherical surface and $R$ is the radius of curvature).

$A$ small object is enclosed in a transparent solid sphere of radius $8 \ cm$. The object is situated at $2 \ cm$ from the centre of the sphere. If its image appears to be at $3.2 \ cm$ from the nearest side,then the refractive index of the material of the sphere is:

An air bubble in a glass sphere $(\mu=1.5)$ is situated at a distance $3\; cm$ from a convex surface of diameter $10\; cm$. At what distance from the surface will the bubble appear (in $; cm$)?

Light from a point source in air falls on a spherical glass surface ($n = 1.5$ and radius of curvature $R = 20 \ cm$). The distance of the light source from the glass surface is $100 \ cm$. Find the image distance. (in $cm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module