- An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the image formation and hence derive the mirror equation $\frac{1}{\text{f}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}.$
- An object is placed 30cm in front of a plano-convex lens with its spherical surface of radius of curvature 20cm. If the refractive index of the material of the lens is 1.5, find the position and nature of the image formed.
CBSE 55-1-3 PAPER SET 2020
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$\triangle\text{ABC}\sim\triangle\text{A}_1\text{B}_1\text{C}$
$\Rightarrow\frac{\text{A}_1\text{B}_1}{\text{AB}}=\frac{\text{A}_1\text{C}}{\text{AC}}=\frac{(+\text{v})+(-\text{R)}}{(-\text{R})-(-\text{u)}}\ ...(1)$
$\triangle\text{ABP}\sim\triangle\text{A}_1\text{B}_1\text{P}$
$\Rightarrow\frac{\text{A}_1\text{B}_1}{\text{AB}}=\frac{\text{A}_1\text{P}}{\text{AP}}=\frac{+\text{v}}{-\text{u}}\ ...(2)$
$(1)=(2)$
$\Rightarrow\frac{\text{v}-\text{R}}{-\text{R}+\text{u}}=\frac{\text{v}}{-\text{u}}$
$\Rightarrow-\text{uv}+\text{uR}=-\text{vR}+\text{uv}$
$\Rightarrow\text{uR}+\text{vR}=2\text{uv}$
$\div\text{ by }\text{uvR}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{2}{\text{R}}$
$\because\text{R}=2\text{F}$
$\therefore\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
- $\frac{1}{\text{f}}=(1.5-1)\Big(\frac{1}{20}-\frac{1}{\infty}\Big)$
$\therefore\text{f}=40\text{cm}$
Now, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{v}}$
$\Rightarrow\text{v}=\frac{\text{fu}}{\text{f}+\text{u}}=\frac{40\times-30}{40-30}$
$\Rightarrow\text{v}=\frac{-40\times30}{10}=-120\text{cm}$
Image is virtual, erect and enlarged in front of lens 120cm away.
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