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  • Please help! A particle moves towards a concave mirror of focal length 30 cm along its axis and with a constant speed of 4 cm/ sec. What is the speed of its image when the particle is at 90 cm from the mirror?

A particle moves towards a concave mirror of focal length 30 cm along its axis and with a constant speed of 4 cm/ sec. What is the speed of its image when the particle is at 90 cm from the mirror?

  • Option 1)

    2 cm/ sec

  • Option 2)

    8 cm/sec

  • Option 3)

    1 cm/sec

  • Option 4)

    4 cm/sec

 

Answers (1)

best_answer

 

Relation between object and image velocity if object is moving along principal axis -

\underset{V_{im}}{\rightarrow} = \frac{-v^{2}}{u^{2}} \underset{V_{om}}{\rightarrow}

\underset{V_{im}}{\rightarrow} = \underset{V_{i}}{\rightarrow} - \underset{V_{m}}{\rightarrow}

\underset{V_{om}}{\rightarrow} = \underset{V_{o}}{\rightarrow} - \underset{V_{m}}{\rightarrow}
 

 

 

 

 

- wherein

\underset{V_{o}}{\rightarrow} = velocity of light

\underset{V_{i}}{\rightarrow} = velocity of image

\underset{V_{m}}{\rightarrow} = velocity of mirror

\underset{V_{om}}{\rightarrow} = velocity of object w.r.t. mirror

\underset{V_{im}}{\rightarrow} = velocity of image w.r.t. mirror

 

 

\frac{dv}{dt}  =   \frac{-v^{2}}{u^{2}}. \frac{du}{dt} ;  

u= -90 f= 30 cm

\frac{1}{v} = \frac{1}{f} -\frac{1}{u}=\frac{-1}{30}+\frac{1}{90} =\frac{-2}{90}

v = -45 cm

\Rightarrow \frac{dv}{dt} = -\left ( \frac{-45}{-90} \right )^{2}\times4 cm/s

= -1 cm/sec  


Option 1)

2 cm/ sec

Incorrect

Option 2)

8 cm/sec

Incorrect

Option 3)

1 cm/sec

Correct

Option 4)

4 cm/sec

Incorrect

Posted by

Aadil

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