
In Ramsden eyepiece, the distance between the eye lens and cross wires is 1.1cm.
What is the focal length of the lenses of the eyepiece?
A. 1.2cm
B. 2.2cm
C. 2.4cm
D. 1.3cm
Answer
487.2k+ views
Hint: Define what is a Ramsden eyepiece. Find the separation of the lenses and the distance of the crosswire from the objective lens in terms of the focal length of the lenses. Find the mathematical expression for the distance between the eye lens and cross wires. Put the values to find the focal length.
Complete answer:
In a Ramsden eyepiece, the eyepiece consists of two plano-convex lenses with the curved part of the lenses facing each other. If the focal length of each of the plano-convex lenses is f, then the lenses are separated by a distance of $\dfrac{2}{3}f$. Also, the distance of the cross-wire from the objective lens is $\dfrac{f}{4}$.
Now, the distance of the cross wire from the eye can be given as the addition of the separation between the plano-convex lenses and the distance of the cross wire from the objective lens.
Given in the question that, in a Ramsden eyepiece the distance between the eye lens and cross wires is 1.1cm.
So, we can write that,
$\begin{align}
& d=\dfrac{2}{3}f+\dfrac{f}{4}=1.1cm \\
& \dfrac{11}{12}f=1.1cm \\
& f=\dfrac{12}{11}\times 1.1cm \\
& f=1.2cm \\
\end{align}$
So, the focal length of the lenses of the Ramsden eyepiece will be 1.2cm each.
The correct option is (A).
Note:
The Ramsden eyepiece is effective in reducing the chromatic aberration of a lens. The minimization of chromatic aberration depends on the spacing of the two lenses so that the dependence on the refractive index is minimum. The Ramsden eyepiece can also reduce the effects of spherical aberration and distortion. It can also remove the problem of coma.
Complete answer:
In a Ramsden eyepiece, the eyepiece consists of two plano-convex lenses with the curved part of the lenses facing each other. If the focal length of each of the plano-convex lenses is f, then the lenses are separated by a distance of $\dfrac{2}{3}f$. Also, the distance of the cross-wire from the objective lens is $\dfrac{f}{4}$.

Now, the distance of the cross wire from the eye can be given as the addition of the separation between the plano-convex lenses and the distance of the cross wire from the objective lens.
Given in the question that, in a Ramsden eyepiece the distance between the eye lens and cross wires is 1.1cm.
So, we can write that,
$\begin{align}
& d=\dfrac{2}{3}f+\dfrac{f}{4}=1.1cm \\
& \dfrac{11}{12}f=1.1cm \\
& f=\dfrac{12}{11}\times 1.1cm \\
& f=1.2cm \\
\end{align}$
So, the focal length of the lenses of the Ramsden eyepiece will be 1.2cm each.
The correct option is (A).
Note:
The Ramsden eyepiece is effective in reducing the chromatic aberration of a lens. The minimization of chromatic aberration depends on the spacing of the two lenses so that the dependence on the refractive index is minimum. The Ramsden eyepiece can also reduce the effects of spherical aberration and distortion. It can also remove the problem of coma.
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