Question

A convex mirror used for rear view on an automobile has a radius of curvature of 3.00 m. if a bus is located at 5.00m from this mirror. Find the position nature and shape of the image

Answer 1

GiveN :

• Radius of Curvature (R) = 3.0 m
• Object Distance (u) = - 5.0 m
• Convex Mirror

To FinD :

• Position, Nature and size of Image produced.

SolutioN :

Find Focal length from Radius of curvature :

$\longrightarrow \sf{R \: = \: 2f} \\ \\ \longrightarrow \sf{f \: = \: \dfrac{R}{2}} \\ \\ \longrightarrow \sf{f \: = \: \dfrac{3}{2}} \\ \\ \longrightarrow \sf{f \: = \: 1.5} \\ \\ \underline{\boxed{\sf{Focal\ length\ =\ 1.5\ m}}}$

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Use Mirror Formula :

$\implies \sf{\dfrac{1}{f}\ =\ \dfrac{1}{v}\ +\ \dfrac{1}{u}} \\ \\ \\ \implies \sf{\dfrac{1}{v}\ =\ \dfrac{1}{f}\ -\ \dfrac{1}{u}} \\ \\ \\ \implies \sf{\dfrac{1}{v}\ =\ \dfrac{1}{1.5}\ -\ \bigg( \dfrac{-1}{5} \bigg) } \\ \\ \\ \implies \sf{\dfrac{1}{v}\ =\ \dfrac{1}{1.5}\ +\ \dfrac{1}{5}} \\ \\ \\ \implies \sf{\dfrac{1}{v}\ =\ \dfrac{10\ +\ 3}{15}} \\ \\ \\ \implies \sf{\dfrac{1}{v}\ =\ \dfrac{13}{15}} \\ \\ \\ \implies \sf{v\ =\ \dfrac{15}{13}} \\ \\ \\ \implies \sf{v\ =\ 1.1538} \\ \\ \\ \implies \sf{v\ \approx\ 1} \\ \\ \\ \underline{\boxed{\sf{Image\ Distance\ =\ 1\ m\ (approx.)}}}$

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Now, use formula for Magnification :

$\implies \sf{m\ =\ \dfrac{-v}{u}} \\ \\ \\ \implies \sf{m\ =\ \dfrac{-1}{-5}} \\ \\ \\ \implies \sf{m\ =\ 0.2}$

• Nature of Image is Virtual and erect
• Size of Image is diminished

Answer 2

Answer:GiveN :

Radius of Curvature (R) = 3.0 m

Object Distance (u) = - 5.0 m

Convex Mirror

To FinD :

Position, Nature and size of Image produced.

SolutioN :

Find Focal length from Radius of curvature :

___________________________

Use Mirror Formula :

___________________________

Now, use formula for Magnification :

Nature of Image is Virtual and erect

Size of Image is diminished

Explanation: