In Ramsdens eye piece the field lens and eye lens have focal lengths f1 and f2 respectively and separated by a distance d, then:

Question

In Ramsdens eye piece the field lens and eye lens have focal lengths f1 and f2 respectively and separated by a distance d, then: (A)  f1=f2(2/3),d=(2/3)f1  (B)  f1=f2,d=f1+f2  (C)  f1=f2,d=(2/3)f  (D)  f1=3f2,d=f1+f2

Tardigrade Admin · Accepted Answer

It consists of two coaxial plano-convex lenses of equal focal lengths f1 and f2 separated by a distance (2/3)f. It means, f1=f2 and d=(2/3)f

Q. In Ramsdens eye piece the field lens and eye lens have focal lengths f1 and f2 respectively and separated by a distance d, then:

$f_{1} = f_{2} \frac{2}{3}, d = \frac{2}{3} f_{1}$

$f_{1} = f_{2}, d = f_{1} + f_{2}$

$f_{1} = f_{2}, d = \frac{2}{3} f$

$f_{1} = 3 f_{2}, d = f_{1} + f_{2}$

It consists of two coaxial plano-convex lenses of equal focal lengths $f_{1}$ and $f_{2}$ separated by a distance $\frac{2}{3} f .$
It means, $f_{1} = f_{2}$ and $d = \frac{2}{3} f$

Q. In Ramsdens eye piece the field lens and eye lens have focal lengths f1 and f2 respectively and separated by a distance d, then:

f1​=f2​32​,d=32​f1​

f1​=f2​,d=f1​+f2​

f1​=f2​,d=32​f

f1​=3f2​,d=f1​+f2​

It consists of two coaxial plano-convex lenses of equal focal lengths f1​ and f2​ separated by a distance 32​f. It means, f1​=f2​ and d=32​f

$f_{1} = f_{2} \frac{2}{3}, d = \frac{2}{3} f_{1}$

$f_{1} = f_{2}, d = f_{1} + f_{2}$

$f_{1} = f_{2}, d = \frac{2}{3} f$

$f_{1} = 3 f_{2}, d = f_{1} + f_{2}$

It consists of two coaxial plano-convex lenses of equal focal lengths $f_{1}$ and $f_{2}$ separated by a distance $\frac{2}{3} f .$
It means, $f_{1} = f_{2}$ and $d = \frac{2}{3} f$