1The answer is \sqrt{a_{1}a_{2}}.
The size of d object is geometric mean of size of images.
A lens is placed between a source of light and wall. It forms images of area a1 and a2 on the wall for two positions. The area of source is
A)\frac{\sqrt{a_{1}a_{2}}}{2} B) \frac{a_{1}a_{2}}{a_{1}+a_{2}} C) \sqrt{a_{1}a_{2}} D) None of these
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6 Answers
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3actually these guys hav remembered the formula ..which is y they got it so fast ........it can be derived ............ Grandmaster
1FIRST OF ALL CHECK DIMENSIONS OF OPTIONS AND THEN PUT a1=a2=a THEN ANSWER MUST BE a
3@harsh AISYE ANSWERS NIKALTE HAIN???[3]
MANMAY DISPLACEMENT METHOD TH0DA PADHNA......
LET IF SOURCE HAS SIDES AS a and b so area of image =ma*mb where m is the magnification
from DISPLACEMENT METHOD for two positions of object m1*m2=1
so here m12ab=a1
m22ab=a2
multiuply these two equations ...
1*a2b2=a1a2
there fore ab=area of source =√a1a2
