Answer
64.8k+ views
Hint in this question, we need to find the relation between the resolving power of the microscope and the refractive index of the glass used. It is given that the objective lens makes an angle $\beta $ with the focus of the microscope. This means that the total aperture angle would be equal to 2 $\beta $.
Complete step by step solution
We know that the minimum separation between to objects for them to be successfully resolved by a microscope is given by the following expression:
${d_{\min }} = \dfrac{{1.22\lambda }}{{2n\sin \theta }}$
Where, n is the refractive index of the material and \[\lambda \] is the wavelength of the light used.
The resolving power is inversely proportional to this minimum separation between the two objects.
\[ \Rightarrow R.P. = \dfrac{1}{{{d_{\min }}}}\]
Now, we will substitute the value of \[{d_{min}}\] into this expression for resolving power,
\[
\Rightarrow R.P. = \dfrac{1}{{\dfrac{{1.22\lambda }}{{2n\sin \theta }}}} \\
\Rightarrow R.P. = \dfrac{{2n\sin \theta }}{{1.22\lambda }} \\
\]
It is given to us that the diameter of the objective lens of the microscope makes an angle ‘ $\beta $ ’ at the focus of the microscope.
Hence, $\theta = 2\beta $ .
On substituting this, the final expression for resolving power of microscope becomes,
\[ \Rightarrow R.P. = \dfrac{{2n\sin 2\beta }}{{1.22\lambda }}\]
From this expression we can deduce that our resolving power of the microscope is directly proportional to \[n\sin 2\beta \]. So, the resolving power increases with increase in \[n\sin 2\beta \] .
Therefore, option (C) is correct.
Note The physical significance that one must know here, is that the resolving power of a microscope is the ability of the microscope to form two separate images for two very closely placed objects.
Complete step by step solution
We know that the minimum separation between to objects for them to be successfully resolved by a microscope is given by the following expression:
${d_{\min }} = \dfrac{{1.22\lambda }}{{2n\sin \theta }}$
Where, n is the refractive index of the material and \[\lambda \] is the wavelength of the light used.
The resolving power is inversely proportional to this minimum separation between the two objects.
\[ \Rightarrow R.P. = \dfrac{1}{{{d_{\min }}}}\]
Now, we will substitute the value of \[{d_{min}}\] into this expression for resolving power,
\[
\Rightarrow R.P. = \dfrac{1}{{\dfrac{{1.22\lambda }}{{2n\sin \theta }}}} \\
\Rightarrow R.P. = \dfrac{{2n\sin \theta }}{{1.22\lambda }} \\
\]
It is given to us that the diameter of the objective lens of the microscope makes an angle ‘ $\beta $ ’ at the focus of the microscope.
Hence, $\theta = 2\beta $ .
On substituting this, the final expression for resolving power of microscope becomes,
\[ \Rightarrow R.P. = \dfrac{{2n\sin 2\beta }}{{1.22\lambda }}\]
From this expression we can deduce that our resolving power of the microscope is directly proportional to \[n\sin 2\beta \]. So, the resolving power increases with increase in \[n\sin 2\beta \] .
Therefore, option (C) is correct.
Note The physical significance that one must know here, is that the resolving power of a microscope is the ability of the microscope to form two separate images for two very closely placed objects.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)