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Klio2033 [76]
3 years ago
8

A scientist designs an experiment in which a light source is pointed at a substance and a light detector is placed on the opposi

te side of the substance.
When the light source is turned on, the scientist observes that no light is detected on the far side of the substance. What is the best interpretation for this observation?

A. The light was either reflected or transmitted by the substance.
B. The light was both absorbed and reflected by the substance.
C. The light was both reflected and transmitted by the substance.
D. The light was either absorbed or reflected by the substance.

Physics
2 answers:
worty [1.4K]3 years ago
6 0

Answer:

option D

Explanation:

When a light falls on a substance, it can reflect, absorb or transmitted by it. It is possible that a combination of above phenomenon occurs.

In the given situation, as no light is detected by the light detector, it means that no light is transmitted by the substance. The light must be either absorbed or bounced back (reflected).

Thus, option D is correct.

Lostsunrise [7]3 years ago
5 0
The answer is D. Either absorbed or reflected. The reason is because if no light is being shown on the other side, the substance is not letting any light pass through. Since we do not know anything else about the substance, we do not know which one of the two it is doing. The scientist would not see any light on the detector if 100% of the light is reflected and the same thing would happen if 100% of the light was absorbed.
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Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as

              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

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          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

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                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

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                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

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