Question 18: Ray Optics

Question

A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be

Accepted Answer

For a compound microscope, the magnifying power is
$$ M = m_o \, m_e $$
where $m_o$ is the magnification produced by the objective and $m_e$ is that of the eyepiece.
Since only the objective is changed, and we want the
same total magnification
, the magnification due to the objective must remain the same.
Step 1: Focal length of the original objective
The objective is a symmetric biconvex lens.
Using lens makerÃ¢ÂÂs formula:
$$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight) $$
For a symmetric biconvex lens,
$$ R_1 = R, \qquad R_2 = -R $$
So,
$$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R} - \left(-\frac{1}{R}ight)ight) = (\mu - 1)\frac{2}{R} $$
Thus,
$$ \frac{1}{f} = \frac{2(\mu-1)}{R} $$
Step 2: Focal length after cutting the lens vertically
When the lens is cut
vertically
, the curvature of the two spherical surfaces does
not change
. Only the aperture is reduced.
So each half still has the
same focal length
as the original lens:
$$ f' = f $$
Hence, the objective lens power remains unchanged.
Step 3: Condition for same magnification with object distance unchanged
For the objective lens,
$$ m_o = \frac{v_o}{u_o} $$
Since the object distance $u_o$ is unchanged, and focal length is also unchanged, by lens formula,
$$ \frac{1}{f} = \frac{1}{v_o} - \frac{1}{u_o} $$
the image distance $v_o$ also remains unchanged.
Therefore, magnification of the objective remains unchanged.
Since the eyepiece is also unchanged, total magnification remains unchanged without changing tube length.
So, ideally,
tube length need not be changed at all
.
But this option is not given. In such questions, sometimes the examiner assumes the lens is cut in a way that one curved surface and one plane surface form a plano-convex lens, and then its focal length changes.
Let us check that.
Step 4: If one half behaves as a plano-convex lens
For a plano-convex lens,
$$ \frac{1}{f'} = (\mu - 1)\left(\frac{1}{R} - 0ight) = \frac{\mu - 1}{R} $$
Comparing with original lens:
$$ \frac{1}{f} = \frac{2(\mu - 1)}{R} $$
So,
$$ f' = 2f $$
Thus focal length of objective becomes double.
Now objective magnification approximately is
$$ m_o \approx \frac{L}{f_o} $$
where $L$ is the tube length.
To keep magnification same:
$$ \frac{L'}{f'} = \frac{L}{f} $$
Substitute $f' = 2f$:
$$ \frac{L'}{2f} = \frac{L}{f} $$
$$ L' = 2L $$
So tube length must be
doubled
.
Final answer
$$ \boxed{	ext{Option A: increased two times}} $$

Answer

increased two times

Answer

increased $\frac{3}{2}$ times

Answer

decreased two times

Answer

decreased $\frac{3}{2}$ times

Ray Optics

Choose the correct answer: