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Andrew [12]
3 years ago
8

. Rohit had only one light source in his room. He was told by his friend that he can make the room more lighted by using mirror.

What type of mirror can be used for the purpose?
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0

Answer:

very smooth and polished glass mirror

Explanation:

The best type of mirror for this would be a very smooth and polished glass mirror. A metal mirror will also work but it would need to be extremely polished. This would allow light to efficiently be reflected across the room, thus effectively adding an "extra" light source to the room. If the mirror is not smooth then the light may become distorted and not reflect properly as it is supposed to. This may defeat the entire purpose of the mirror for this scenario.

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You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.
Neporo4naja [7]

Answer:

The distance is r_2  =  0.24 \  m

Explanation:

From the question we are told that

       The  distance from the conversation is r_1    =  24.0 \ m

       The  intensity of  the sound at your position is  \beta _1 =  40 dB

        The  intensity at the sound at the new position is  \beta_2 =  80.0dB

Generally the intensity in  decibel is  is mathematically represented as

      \beta  =  10dB log_{10}[\frac{d}{d_o} ]

The intensity is  also mathematically represented as

      d =  \frac{P}{A}

So

    \beta  =  10dB *  log_{10}[\frac{P}{A* d_o} ]

=>   \frac{\beta}{10}  =  log_{10} [\frac{P}{A (l_o)} ]

From the logarithm definition

=>    \frac{P}{A  *  d_o}  =  10^{\frac{\beta}{10} }

=>      P =  A (d_o ) [10^{\frac{\beta }{ 10} } ]

Here P is the power of the sound wave

 and  A is the cross-sectional area of the sound wave  which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

               P_1 =  A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]

Now the power of the sound wave at the second  position is mathematically represented as

               P_2 =  A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

Generally  power of the wave is constant at both positions  so  

    A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]  = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

      4 \pi r_1 ^2   [10^{\frac{\beta_1 }{ 10} } ]  = 4 \pi r_2 ^2   [10^{\frac{\beta_2 }{ 10} } ]

        r_2 =  \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}

       substituting value

        r_2 =   \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}

        r_2  =  0.24 \  m

     

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Answer:

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Answer:

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Explanation:

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Explanation:

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Explanation:

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