Chapter 23, Problem 104QAP

To determine

(a)

The three smallest angles on the either side of central maximum at which no light will reach the cat's retina and it is assumed the eyeball of cat is filled with air.

Answer to Problem 104QAP

1^st smallest angle = ${\theta }_1\text{=1}\text{.1}\times {\text{10}}^{-3}rad$

2^nd smallest angle = ${\theta }_2\text{=2}\text{.2}\times {\text{10}}^{-3}rad$

3^rd smallest angle = ${\theta }_3\text{=3}\text{.3}\times {\text{10}}^{-3}rad$

Explanation of Solution

Given:

Wavelength of light = $\lambda =550nm=5.50\times {10}^{-7}m$

Width of slit = $w=0.500mm=0.5\times {10}^{-3}m$

Formula used:

Following equation satisfy for the diffraction of single slit,

  $w\sin \theta =m\lambda$

Calculation:

Since, w is very small to $\lambda$, so, $\theta \text{would be very small}\text{.}$

In this situation, we can apply the following approximation,

  $\sin \theta \approx \theta =\frac{m\lambda }{w}$

Now, let's plug all values in above equation,

  $\begin{array}{l}\theta =\frac{m\lambda }{w}=\frac{m(5.50\times {10}^{-7}m)}{0.5\times {10}^{-3}m}=m\left(1.1\times {10}^{-3}rad\right)\\ \text{now, when m = 1,}\\ {\theta }_1\text{=1}\text{.1}\times {\text{10}}^{-3}rad\\ \text{when m = 2,}\\ {\theta }_2\text{=2}\text{.2}\times {\text{10}}^{-3}rad\\ \text{When m = 3,}\\ {\theta }_3\text{=3}\text{.3}\times {\text{10}}^{-3}rad\end{array}$

Conclusion:

So, the three smallest angles on the either side of central maximum at which no light will reach the cat's retina and medium inside the eye ball is air are $\text{1}\text{.1}\times {\text{10}}^{-3}rad,2.2\times {\text{10}}^{-3}rad,3.3\times {\text{10}}^{-3}rad.$

To determine

(b)

The three smallest angles on the either side of central maximum at which no light will reach the cat's retina and it is assumed the eyeball of cat is filled with a liquid having refractive index 1.4.

Answer to Problem 104QAP

1^st smallest angle = ${\theta }_1\text{=7}\text{.86}\times {\text{10}}^{-4}rad$

2^nd smallest angle = ${\theta }_2\text{=1}\text{.57}\times {\text{10}}^{-3}rad$

3^rd smallest angle = ${\theta }_3\text{=2}\text{.36}\times {\text{10}}^{-3}rad$

Explanation of Solution

Given:

Refractive index fluid that is in the eyeball,

  $n=1.4$

Wavelength of light in air = $\lambda =550nm=5.50\times {10}^{-7}m$

Width of slit = $w=0.500mm=0.5\times {10}^{-3}m$

Formula used:

Wavelength of light in medium is written as,

  $\begin{array}{l}\lambda \text{'}=\frac{\lambda }{n}\\ \lambda \text{'}=\text{wavelength of light in medium}\\ \lambda \text{=wavelength of light in air}\\ \text{n}\text{=}\text{refractive index}\end{array}$

Calculation:

Wavelength of light in fluid,

  $\lambda \text{'}=\frac{\lambda }{n}=\frac{5.50\times {10}^{-7}m}{1.4}=3.929\times {10}^{-7}m$

Now, from above formula,

  $\begin{array}{l}w\sin \theta =m\lambda \text{'}\\ \sin \theta \approx \theta =\frac{m\lambda \text{'}}{w}\end{array}$

Let's plug all values,

  $\begin{array}{l}\theta =\frac{m\lambda \text{'}}{w}=\frac{m(3.929\times {10}^{-7}m)}{0.5\times {10}^{-3}m}\\ \theta =m(7.857\times {10}^{-4}rad)\\ \text{now, if m = 1,}\\ {\theta }_1=7.857\times {10}^{-4}rad\\ \text{if m = 2,}\\ {\theta }_2=1.57\times {10}^{-3}rad\\ \text{if m = 3,}\\ {\theta }_3=2.36\times {10}^{-3}rad\end{array}$

Conclusion:

Thus, the three smallest angles on either side of retina of cat's eye when eyeball is filled with fluid having refractive index $n=1.4$ are $\text{7}\text{.86}\times {\text{10}}^{-4}rad,1.57\times {\text{10}}^{-3}rad\text{and 2}\text{.36}\times {\text{10}}^{-3}rad.$

To determine

(c)

The explanation of the order of fringes due to diffraction of light by cat's eye

Answer to Problem 104QAP

Cat cannot distinguish the location of fringe on the retina.

Explanation of Solution

From above discussion, we can see that angle of diffraction of light on either side of central bright spot of the retina is very small. So, location of bright and dark fringes would be very close to each other. Cat's eye has no ability to perceive that order of fringes. So, cat would perceive it as continuous bright light not as alternate dark and bright that occurs due to diffraction.

Conclusion:

Thus, cat would not see that alternate dark and bright fringes.