at the first lens
1/v + 0 = 1/220*10-3
1/v= 1000/220
v =220mm
1/v - 1/200mm = 1/140mm
1/v = 1000/140 +1000/200
1/v = 200*1000 + 140*1000/140*200
1/v = 200 + 140 /28
1/v = 340/28
v = 28/340
v = 0.082m ahead of the second
A Cassegrain telescope uses two mirrors 20 mm apart. radius of curvature of the mirrors 220mm and 140 mm.when object at infinity..final image will be formed at...???
at the first lens
1/v + 0 = 1/220*10-3
1/v= 1000/220
v =220mm
1/v - 1/200mm = 1/140mm
1/v = 1000/140 +1000/200
1/v = 200*1000 + 140*1000/140*200
1/v = 200 + 140 /28
1/v = 340/28
v = 28/340
v = 0.082m ahead of the second
sorry, virang....didn't see ur post
but i got the answer as 315 mm in front of the convex mirror
Virang, 220 mm is the RADIUS OF CURVATURE of the mirror.......
in the first mirror, the rays are from infnity so they converge at the focus.... so v= -110 mm.
But the convex mirror is 20 mm in front of the concave mirror. So the rays hit the mirror as if they are coming from an object 90 mm behind it. Also f for convex lens is 70mm. So
\frac{1}{70}=\frac{1}{90}+\frac{1}{v}
So v = 315 mm
So it is formed 315 - 20 (= 295 mm) behind the first mirror.
I may have made a mistake in the sign conventions or calculations, but i'm sure of the method....