final image formed at what distance?
Asked by abinash.gupta003 | 19th Dec, 2017, 12:01: AM
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.
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
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