SATHEE: Optics 1 Question 14

Optics 1 Question 14

14. Image of an object approaching a convex mirror of radius of curvature $20 m$ along its optical axis is observed to move from $\frac{25}{3} m$ to $\frac{50}{7} m$ in $30 s$ . What is the speed of the object in $\underset{(2010)}{k m h^{- 1} ?}$

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Answer:

Correct Answer: 14. 3

Solution:

Using mirror formula twice,

$\begin{aligned} \frac{1}{+ 25 / 3} + \frac{1}{- u_{1}} = \frac{1}{+ 10} \\ or \frac{1}{u_{1}} = \frac{3}{25} - \frac{1}{10} \\ or u_{1} = 50 m and \frac{1}{(+ 50 / 7)} + \frac{1}{- u_{2}} = \frac{1}{+ 10} \\ ∴ \frac{1}{u_{2}} = \frac{7}{50} - \frac{1}{10} or u_{2} = 25 m \end{aligned}$

Speed of object $= \frac{u_{1} - u_{2}}{time}$

$= \frac{25}{30} m s^{- 1} = 3 k m h^{- 1}$

$∴$ Answer is 3 .

Optics 1 Question 14

14. Image of an object approaching a convex mirror of radius of curvature $20 m$ along its optical axis is observed to move from $\frac{25}{3} m$ to $\frac{50}{7} m$ in $30 s$ . What is the speed of the object in $\underset{(2010)}{k m h^{- 1} ?}$

Answer:

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Optics 1 Question 14

14. Image of an object approaching a convex mirror of radius of curvature 20m along its optical axis is observed to move from 253m to 507m in 30s. What is the speed of the object in kmh−1 ? (2010)

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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14. Image of an object approaching a convex mirror of radius of curvature $20 m$ along its optical axis is observed to move from $\frac{25}{3} m$ to $\frac{50}{7} m$ in $30 s$ . What is the speed of the object in $\underset{(2010)}{k m h^{- 1} ?}$