final image formed at what distance?

Asked by abinash.gupta003 | 19th Dec, 2017, 12:01: AM

Expert Answer:

Refer the above diagram for the following explanation. Let the object is placed at S. Due to refraction in the liquid, the object distance above the liquid surface which is 10 cm,  is increased as shown in figure. 
Increased distance of S from liquid surface = 10 × (4/3) = 40/3.
Hence the object distance seen by the mirror due to refraction is (40/3) + (5/3) = 15 cm
Let us get the image distance from the equation that involves the object distance, image distance and focal length.
begin mathsize 12px style 1 over f space equals space 1 over u plus 1 over v space
1 over v equals space 1 over f minus 1 over u equals space 1 over 10 minus 1 over 15 semicolon space t h i s space g i v e s space v space equals space 30 space c m space semicolon end style
Hence the distance from mirror to the image point I' is 30 cm. But this includes the enhanced distance of I' above liquid surface due to refraction.
Hence the actual distance of image I above the liquid surface is [ 30 - (5/3) ]× (3/4) = 85/4 cm

Answered by  | 19th Dec, 2017, 03:06: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.