Optics 6 Question 6
7. The focal lengths of the objective and the eyepiece of a compound microscope are $2.0 cm$ and $3.0 cm$ respectively. The distance between the objective and the eyepiece is $15.0 cm$. The final image formed by the eyepiece is at infinity. The two lenses are thin. The distance in $cm$ of the object and the image produced by the objective, measured from the objective lens, are respectively
(a) 2.4 and 12.0
(b) 2.4 and 15.0
(c) 2.0 and 12.0
(d) 2.0 and 3.0
$(1995,2 M)$
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Solution:
- Let intensity of each wave is $I _0$.
Then, intensity at the centre of bright fringe will be $4 I _0$.
Given, path difference, $\Delta x=\lambda / 8$
$\therefore$ Phase difference, $\varphi=\Delta x \times \frac{2 \pi}{\lambda}$
$$ \Rightarrow \quad \varphi=\frac{\lambda}{8} \times \frac{2 \pi}{\lambda} \text { or } \varphi=\pi / 4 $$
Intensity of light at this point,
$$ I^{\prime}=I _0+I _0+2 I _0 \cos (\pi / 4)=2 I _0+\sqrt{2} I _0=3.41 I _0 $$
Now,
$$ \frac{I^{\prime}}{4 I _0}=\frac{3.41}{4}=0.85 $$