Question

Narrow, parallel, glowing gas filled tubes in a variety of colors from block letters spell out the name of a nightclub. Adjacent tubes are all 2.80 cm apart. The tubes forming one letter are filled with neon and radiate predominantly red light with a wavelength of 640 nm. For another letter, the tubes emit predominantly blue light at 440 nm. The pupil of a dark adapted viewer’s eye is 5.20 mm in diameter. If she is in a certain range of distances away, the viewer can resolve the separate tubes of one color but not the other. The viewer’s distance must be in what range for her to resolve the tubes of only one of these two colors?

Answer 1

In this question we have to find the range of distances for which she can resolve the tubes of one colour only. For this we are going to use the formula of angular separation in terms of wavelength and diameter of aperture.

Complete step by step answer:
When the pupil of the eye is widely open, it appears that the resolving power of the human eye is limited by the coarseness of light sensors on the retina of light. We will use Rayleigh’s criteria to find out how good our vision of eyes might be. We will use following formula of angular separation,
${\theta _{\min }} = \dfrac{d}{L}$
Or ${\theta _{\min }} = 1.22\dfrac{\lambda }{D}$
By this we get following equation
$\dfrac{d}{L} = 1.22\dfrac{\lambda }{D}$
It implies that the formula of L will be as follows:
$L = \dfrac{{Dd}}{{1.22\lambda }}$
Where,
${\theta _{\min }}$ is the smallest angular separation of two objects.
$d$ is the separation two objects
L is the maximum distance of the aperture from two objects at which they can be resolved.
D is the diameter of the aperture
$\lambda$ is wavelength of light
Now, we will put the values of variables given in the question
$L = \dfrac{{Dd}}{{1.22\lambda }}$
$L = \dfrac{{5.20 \times {{10}^{ - 3}} \times 2.80 \times {{10}^{ - 2}}}}{{1.22\lambda }}$
$L = \dfrac{{14.56 \times {{10}^{ - 5}}}}{{1.22\lambda }}$
And hence on solving,we have
$L = \dfrac{{1.193 \times {{10}^{ - 4}}}}{\lambda }$
Now we will find the value of L at two given values of wavelength
At $\lambda = 640nm$
$L = \dfrac{{1.193 \times {{10}^{ - 4}}}}{{640 \times {{10}^{ - 9}}}}$
$L = 0.00186 \times {10^5}m$
Again on doing the simplification,we have $L = 186m$
At $\lambda = 440nm$
$L = \dfrac{{1.193 \times {{10}^{ - 4}}}}{{440 \times {{10}^{ - 9}}}}$
And ence finally,we have
$L = 0.00271 \times {10^5}m$
$L = 271m$
Result- The viewer with the assumed diffraction limited vision can resolve tubes of blue light in the range $L = 186m$ to $L = 271m$, but she cannot resolve adjacent tubes of red in this range.

Note:
In this question we have used two formulae of specific heat. So, the knowledge of these formulae is a must and we should be careful while doing the calculation of the problems. One more thing that we must be careful about is the units of parameters. All the units must be in the same system.
In this question it was quite easy that we just had to apply formula directly but there might be a case when we have to be careful when using the formulae because we might have to use some theoretical aspects before applying the relevant formula. By following these steps we can find the correct answer to the problem.