(A) For refraction at a spherical surface,the formula is: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,$\mu_1 = 1$ (air) and $\m

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(A) For refraction at a spherical surface,the formula is: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,$\mu_1 = 1$ (air) and $\mu_2 = 1.5$ (glass).According to the sign convention,the object distance $u = -PO = -x$ and the image distance $v = PI = x$,where $x$ is the distance $PO$.The radius of curvature $R$ is positive because the centre of curvature is in the glass medium.Substituting these values into the formula:$\frac{1.5}{x} - \frac{1}{-x} = \frac{1.5 - 1}{R}$$\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R}$$\frac{2.5}{x} = \frac{0.5}{R}$$\frac{5}{2x} = \frac{1}{2R}$$x = 5R$.Therefore,the distance $PO$ is $5R$.

Class 12 Physics Ray Optics and Optical Instruments Refraction Through Single Curved Surface

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$A$ spherical surface of radius of curvature $R$ separates air from glass (refractive index $= 1.5$). The centre of curvature is in the glass medium. $A$ point object $O$ is placed in air on the optic axis of the surface,so that its real image is formed at $I$ inside the glass. The line $OI$ intersects the spherical surface at $P$ and $PO = PI$. The distance $PO$ is equal to: (in $R$)

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A
$5$
B
$3$
C
$2$
D
$1.5$

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Ray Optics and Optical Instruments

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Refraction Through Single Curved Surface

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An air bubble in a glass sphere having a $4 \,cm$ diameter appears $1 \,cm$ from the surface nearest to the eye when looked at along the diameter. If the refractive index of glass is $_a\mu_g = 1.5$,the actual distance of the bubble from the refracting surface is.....$cm$.

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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.

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In a thin spherical fish bowl of radius $10 \, cm$ filled with water of refractive index $\mu = \frac{4}{3}$,there is a small fish at a distance of $4 \, cm$ from the centre $C$ as shown in the figure. Where will the image of the fish appear if seen from $E$ (in $, cm$)?

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The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea $(7.8 \, mm)$. This surface separates two media of refractive indices $1$ and $1.34$. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus in $cm$.

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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

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An air bubble in a glass sphere having a $4 \,cm$ diameter appears $1 \,cm$ from the surface nearest to the eye when looked at along the diameter. If the refractive index of glass is $_a\mu_g = 1.5$,the actual distance of the bubble from the refracting surface is.....$cm$.

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.

In a thin spherical fish bowl of radius $10 \, cm$ filled with water of refractive index $\mu = \frac{4}{3}$,there is a small fish at a distance of $4 \, cm$ from the centre $C$ as shown in the figure. Where will the image of the fish appear if seen from $E$ (in $, cm$)?

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

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