Question 50: Wave Optics

Question

The diameter of the objective lens of microscope makes an angle $$\beta $$ at the focus of the microscope. Further, the medium between the object and the lens is an oil of refractive index n. Then the resolving power of the microscope.

Accepted Answer

The smallest detail a microscope can see (its resolution limit) is given by the formula:
$$ d_{	ext{min}} \approx \frac{\lambda}{2 	ext{N.A.}} $$
Here,
N.A.
stands for numerical aperture. The formula for numerical aperture is:
$$ 	ext{N.A.} = n \sin \alpha $$
where
n
is the refractive index of the oil (the medium), and
ÃÂ±
is half the angle made by the cone of light rays coming from the specimen.
The resolving power (RP) of the microscope tells us how well it can see small details. It is the inverse of the resolution limit:
$$ 	ext{RP} = \frac{1}{d_{	ext{min}}} \propto n \sin \alpha $$
Now, the angle subtended by the lens diameter at the focal point is
ÃÂ²
. From geometry, we see that
ÃÂ±
(half the cone angle) is related to
ÃÂ²
by:
$$ \sin \alpha = \sin 2\beta $$
This means the resolving power becomes:
$$ 	ext{RP} \propto n \sin 2\beta $$
So, the microscope will see finer details as the value of $ n \sin 2\beta $ increases. Answer: C.

Answer

Increases with decreasing value of n

Answer

Increases with decreasing value of $$\beta $$

Answer

Increases with increasing value of n sin 2$$\beta $$

Answer

Increases with increasing value of $${1 \over {n\sin 2\beta }}$$

Wave Optics

The diameter of the objective lens of microscope makes an angle $\beta $ at the focus of the microscope. Further, the medium between the object and the lens is an oil of refractive index n. Then the resolving power of the microscope.

Choose the correct answer: