Guest

Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d ( where d centres are displaced by a vertical separation ? as shown in the figure. Taking the origin of coordinates, O, at the centre of the first lens, the x and y coordinates of the focal point of the lens system, for a parallel beam of rays coming from the left, are given by ( a ) x = 1 2 1 2 f f f f + , y = ? ( b ) x = d ( d ) 1 2 1 2 f f - f f + + , y = 1 2 2 ? f + f ( c ) x = d d ( d ) 1 2 1 2 1 - - f f f f f + + , y = d ? ( d ) 1 2 1 - - f f f + ( d ) x = d d ( d ) 1 2 1 2 1 - - f f f f f + + , y = 0

Two thin convex lenses of focal
lengths f1 and f2 are separated
by a horizontal distance d ( where
d < f1 and d < f2 ), and their
centres are displaced by a vertical
separation ? as shown in the
figure. Taking the origin of
coordinates, O, at the centre of
the first lens, the x and y
coordinates of the focal point of the lens system, for a parallel beam of rays coming
from the left, are given by
( a ) x =
1 2
1 2
f f
f f
+
, y = ? ( b ) x =
d
( d )
1 2
1 2
f f -
f f
+
+
, y =
1 2
2
?
f + f
( c ) x =
d
d ( d )
1 2
1 2 1
-
-
f f
f f f
+
+
, y =
d
? ( d )
1 2
1
-
-
f f
f
+
( d ) x =
d
d ( d )
1 2
1 2 1
-
-
f f
f f f
+
+
, y = 0

Grade:12

1 Answers

Aman Bansal
592 Points
12 years ago

Dear Rajen,

he answer is : x=(f1f2+d(f1-d))/(f1+f2-d) , y=r(f1-d)/(f1+f2-d)

Best Of luck

Plz Approve the answer...!!!!

Cracking IIT just got more exciting,It’s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and winexciting gifts by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple click here to download the toolbar….
So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free