Question

The ratio of resolving powers of an optical microscope for two wavelengths ${\lambda _1} = 4000{A^0}$and ${\lambda _2} = 6000{A^0}$is:
$\left( a \right)$ $16:18$
$\left( b \right)$ $8:27$
$\left( c \right)$ $9:4$
$\left( d \right)$ $3:2$

Answer 1

Hints: So the resolving power of a lens can be known by its power to differentiate between the objects. Hence, as larger will be the power then smaller will be the minimum distance between the lines that can be differ from one another. The larger the N.A., the upper the resolving power.
Formula used:
Resolving power $ \prec \dfrac{1}{{{\lambda _0}}}$
The above relation shows that the wavelength is inversely proportional to the power of the microscope.

Complete step by step solution: we have to find the ratio resolving power, so for this, we have two known values that are known as the wavelength which is given to us.
So by using this formula
Resolving power $ \prec \dfrac{1}{{{\lambda _0}}}$
We can write it as,
$ \Rightarrow \dfrac{{R{P_1}}}{{R{P_2}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}$
Putting the values of the wavelength, we get
$ \Rightarrow \dfrac{{6000{A^0}}}{{4000{A^0}}}$
On solving, we will get the required ratio,
$ \Rightarrow \dfrac{3}{2}$
Hence, $3:2$ is the ratio of resolving power.
Additional information: For microscopes, the resolving power is that the inverse of the gap between two objects that may be simply resolved. This can be given by the known abbe’s criterion given by Ernst Abbe $1873$ as
$ \Rightarrow \vartriangle d = \dfrac{\lambda }{{2n\sin \theta }}$
 Resolving power resolving power $\dfrac{1}{{\vartriangle d}} = \dfrac{{2n\sin \theta }}{\lambda }$
Where; $n$ makes the separation in the refractive index of the medium between aperture and the object. Note that to realize high-resolution $n\sin \theta $ should be massive. This can be called the Numerical aperture.

Notes: So for a better resolution we can do these things. $\sin \theta $ Should be massive. Specimens are kept as close as possible to the target so that we achieve the best out of it. A higher refractive index medium should be used. So to extend the refractive index of microscopes we can use the oil immersion microscopes. This suggests that sometimes organelles, viruses, and proteins can't be imaged.