Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

solve

qsnImg
Asked by adjacentcalliber10 10th May 2019, 11:37 AM
Answered by Expert
Answer:
we have miiror equation (1/u) + (1/v) = (1/f)  ...................(1)
 
where u is mirror-to-object distance ( sign -ve for concave mirror),  v is mirror-to-image distance and f is focal length ( sign -ve for concave mirror )
 
Initial image distance is obtained from eqn.(1) with apllying appropriate sign convention
 
(-2/f) + (1/v ) = (-1/f)     or v  = f  , image is formed behind the mirror, virtual
 
initial magnifiacation m = -v/u  = -f/(-f/2)   = 2   .................(2)
 
after moving the object f/4 away from mirror, image  distance is obtained from eqn.(1) :- (-4/3f) + (1/v) = (-1/f)   or  v = 3f
 
magnification m = -v/u  = -3f/(-3f/4)  = 4  ...................(3)
 
Ratio of magnifications = 4/2 = 2
Answered by Expert 10th May 2019, 12:45 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp