From Fig.33-2, approximate the (a) smaller (b) larger wavelength at which the eye of a standard observer has half the eye’s maximum sensitivity. What are the (c) wavelength, (d) frequency, and (e) period of the light at which the eye is the most sensitive?

Speed of light, $c=3\times {10}^8\text{m/s}$.

The given figure 33-2 shows the relative sensitivity of the average human eye to electromagnetic waves at different wavelengths. This portion of the electromagnetic spectrum to which the eye is sensitive is called visible light. Its range is 400 nm to 700 nm. Using the given figure, we can find the required wavelengths.

From the graph given in fig 33-2, we can observe that the smaller wavelength, at which the eye of a standard observer has half the eye’s maximum sensitivity, is approximately 515 nm.

Thus, the smaller approximate wavelength at which the eye of a standard observer has half the eye’s maximum sensitivity is 515 nm.

From the graph given in fig 33-2, we can observe that the larger wavelength, at which the eye of a standard observer has half the eye’s maximum sensitivity, is approximately 610 nm.

Thus, the larger approximate wavelength at which the eye of a standard observer has half the eye’s maximum sensitivity is 610 nm.

From the figure, the wavelength at which the eye is the most sensitive is about 555 nm.

Thus, the wavelength of the light at which the eye most sensitive is 555 nm.

To find the frequency, we have from c), $\lambda =555\times {10}^{-9}\mathrm{m}$. Therefore,

$f=\frac{c}{\lambda }$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}f& =& \frac{3\times {10}^8}{555\times {10}^{-9}}\\ & =& 5.41\times {10}^{14}\mathrm{Hz}\end{array}$

Thus, the wavelength of the light at which the eye the most sensitive is $5.41\times {10}^{14}\mathrm{Hz}$.

The time period can be calculated as,

$T=\frac{1}{f}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}T& =& \frac{1}{5.41\times {10}^{14}}\\ & =& 1.85\times {10}^{-15}\mathrm{s}\\ & & \end{array}$

Thus, the time period is $1.85\times {10}^{-15}\mathrm{s}$.