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 sarveshvibrantacademy 7th May 2019, 12:01 PM
Answered by Expert
Answer:
Let D be the distance of the object from the pole
Then image formed is at a distance D-10 from the pole
 
As we know 
m equals fraction numerator negative v over denominator u end fraction equals h subscript I over h subscript o
Therefore
 
fraction numerator negative open parentheses D minus 10 close parentheses over denominator D end fraction equals fraction numerator negative 1 over denominator 2 end fraction
2 D minus 20 equals D
D equals 20 c m
D minus 10 equals 20 minus 10 equals 10 c m
 
There
object distance u=-20
Image distance v=-10
 
fraction numerator 1 over denominator negative 10 end fraction plus fraction numerator 1 over denominator negative 20 end fraction equals 1 over f
f equals fraction numerator negative 20 over denominator 3 end fraction
R equals 2 f
T h e r e f o r e
R equals negative 40 divided by 3
D i s tan c e space b e t w e e n space r a d i u s space o f space c u r v a t u r e space a n d space o b j e c t space i s space g i v e n space b y
equals 20 minus 40 over 3
equals 20 over 3 c m
The distance between the pole and object is object distance which is equal to 20cm
Answered by Expert 7th May 2019, 12:39 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