Optics 6 Question 5
6. In a compound microscope, the intermediate image is
$(2000,2 M)$
(a) virtual, erect and magnified
(b) real, erect and magnified
(c) real, inverted and magnified
(d) virtual, erect and reduced
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Answer:
Correct Answer: 6. (a)
Solution:
- Here, wavelength of light used $(\lambda)=5303 \AA$.
Distance between two slit $(d)=19.44 \mu-m$
Width of single slit $(a)=4.05 \mu m$
Here, angular width between first and second diffraction minima
$$ \theta \simeq \frac{\lambda}{a} $$
and angular width of a fringe due to double slit is
$$ \theta^{\prime}=\frac{\lambda}{d} $$
$\therefore$ Number of fringes between first and second diffraction minima, $n=\frac{\theta}{\theta^{\prime}}=\frac{\frac{\lambda}{a}}{\frac{\lambda}{d}}=\frac{d}{a}=\frac{19.44}{4.05}=4.81$ or $n \simeq 5$
$\therefore 5$ interfering bright fringes lie between first and second diffraction minima.
