An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
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$h _0=+5.0 cm, u =-20 cm$,,$\text{f}=\frac{\text{R}}{2}=+15 \ \text{cm}$
Using mirror formula, $\frac{1}{\text{f}}=\frac{1}{\text{v}}+\frac{1}{\text{u}}$, we get$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{1}{15}-\frac{1}{-20}$
$=\frac{20+15}{300}=\frac{35}{300}$
$\text{v}=\frac{300}{35}=\frac{60}{7}=8.57 \ \text{cm}$.
Using $\text{m}=\frac{\text{h}_1}{\text{h}_0}=-\frac{\text{v}}{\text{u}}$, we get$\text{h}_1=-5\times\frac{8.57}{-20}=2.16 \ \text{cm}$
Since v is +ve, the image is virtual. since $h_1$ = 2.16 cm < 5.0 cm, the image is diminished.
Using mirror formula, $\frac{1}{\text{f}}=\frac{1}{\text{v}}+\frac{1}{\text{u}}$, we get$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{1}{15}-\frac{1}{-20}$
$=\frac{20+15}{300}=\frac{35}{300}$
$\text{v}=\frac{300}{35}=\frac{60}{7}=8.57 \ \text{cm}$.
Using $\text{m}=\frac{\text{h}_1}{\text{h}_0}=-\frac{\text{v}}{\text{u}}$, we get$\text{h}_1=-5\times\frac{8.57}{-20}=2.16 \ \text{cm}$
Since v is +ve, the image is virtual. since $h_1$ = 2.16 cm < 5.0 cm, the image is diminished.
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