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Zarrin [17]
3 years ago
10

An experiment is designed to test what color of light will activate a photoelectric cell the best. The photocell is set in a cir

cuit that "clicks" in response to current. The faster the current, the more clicks per minute. In this experiment, the number of clicks in one minute is recorded for each color of light shining on the photocell. To change the color of light, a different color of cellophane is placed over the same flashlight and the flashlight is then located a specific distance from the photocell. Which of the following would be the best hypothesis for this experiment? Red light will activate the photocell to the highest degree. If photoelectric cells respond better to short wavelengths of light, the red light should activate it the least and blue the most. Photoelectric cells are affected by different wavelengths of light.
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
8 0
I believe it would be the 2nd option.
"<span>If photoelectric cells respond better to short wavelengths of light,the red light should activate it the least and blue the most."
Stated above in the question.
</span><span>
</span>
Natalija [7]3 years ago
3 0

A photoelectric cell is an electronic device which is used to convert light energy into electric energy.The operation of this device is based on photoelectric effect.

Light of suitable frequency i.e greater or equal to threshold frequency will fall on the cathode maintained at negative potential.The electron emission will take place and these electrons are drifted towards the anode which is at positive potential.

Here,only those radiations will be capable of emitting electrons irrespective of surface barrier of metals whose energy is greater than the work function.

We know that the radiation having long wavelength has least energy as energy and wavelength are inversely proportional to each other.

Mathematically\ energy\ E=\frac{hc}{\lambda}

Here h is the Planck's constant,c is the velocity of light.

Here we have been given red light and blue light.

In the visible spectrum of radiation, the red light has longer wavelength than all other colors of light.Hence blue light has more energy as it's wavelength is less as compared to red light.

Hence, the blue light will activate the most and red the least.

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Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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