Chapter 36, Problem 7PQ

(a)

To determine

The limiting resolution of the microscope.

Answer to Problem 7PQ

The limiting resolution of the microscope is $8.13\times {10}^{-5}\text{rad}$.

Explanation of Solution

Write the expression for the limit of resolution of the optical system.

    ${\theta }_{\min }=\frac{1.22\lambda }{d}$                                                                                                     (I)

Here, $\lambda$ is the wave length of radio waves and $d$ is the diameter of the telescope and $n$ is the refractive index of medium.

Conclusion:

Substitute $500\text{nm}$ for $\lambda$ and $7.5\text{mm}$ for $d$ in equation (I).

    $\begin{array}{c}{\theta }_{\min }=\frac{1.22\left(500\text{nm}\right)}{7.5\text{mm}}\\ =\frac{1.22\left(500\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{7.5\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =8.13\times {10}^{-5}\text{rad}\end{array}$

Therefore, the limiting resolution of the microscope is $8.13\times {10}^{-5}\text{rad}$.

(b)

To determine

The wavelength that will produce the smallest angle of resolution.

Answer to Problem 7PQ

The smallest angle of resolution is produced by violet light of $400\text{nm}$.

Explanation of Solution

The range of the visible light is from red $\left(\lambda =700\text{nm}\right)$ to violet $\left(\lambda =400\text{nm}\right)$.

From the equation (I), the angle of resolution is proportional to the wavelength of light used. So the shortest wavelength is of violet light which corresponds to the smallest angle of resolution.

Conclusion:

Therefore, the smallest angle of resolution is produced by violet light of $400\text{nm}$ wavelength.

(c)

To determine

The angular resolution of the microscope.

Answer to Problem 7PQ

The angular resolution of the microscope is $6.51\times {10}^{-5}\text{rad}$.

Explanation of Solution

Conclusion:

Substitute $400\text{nm}$ for $\lambda$ and $7.5\text{mm}$ for $d$ in equation (I).

    $\begin{array}{c}{\theta }_{\min }=\frac{1.22\left(400\text{nm}\right)}{7.5\text{mm}}\\ =\frac{1.22\left(400\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{7.5\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =6.51\times {10}^{-5}\text{rad}\end{array}$

Therefore, the angular resolution of the microscope is $6.51\times {10}^{-5}\text{rad}$.

(d)

To determine

The angular resolution of the microscope if the microscope is inserted in water.

Answer to Problem 7PQ

The angular resolution of the microscope is $4.89\times {10}^{-5}\text{rad}$.

Explanation of Solution

Conclusion:

Substitute $400\text{nm}$ for $\lambda$, $1.33$ for $n$ and $7.5\text{mm}$ for $d$ in equation (I).

    $\begin{array}{c}{\theta }_{\min }=\frac{1.22\left(400\text{nm}\right)}{7.5\text{mm}}\\ =\frac{1.22\left(400\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{\left(1.33\right)7.5\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =4.89\times {10}^{-5}\text{rad}\end{array}$

Therefore, the angular resolution of the microscope is $4.89\times {10}^{-5}\text{rad}$.