Science, asked by XxCelestialStarXx, 1 month ago

A Convex lens of focal length 25cm and a concave lens of focal length 10cm are placed in close contact with one another.
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a) What is the power of Combination
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b) What is the focal length of Combination​

Answers

Answered by Anonymous
48

\boxed{ \underline{ \underline{\footnotesize\tt{ Solution:-}}}} \:  \red♡

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\footnotesize\tt{1) \: For  \: Convex \:  lens}

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\footnotesize\tt{Focal \:  length(f) = 25cm}

\footnotesize\tt{Focal \:  length(f) } \tt{ \: =  \frac{25}{100} m}

\footnotesize\tt{Focal \:  length(f) } \tt{ \: = 0.25m}

\boxed{ \underline{ \underline{\footnotesize\tt{ Focal \:  length(f) } \tt{ \: = 0.25m}}}} \:  \red♡

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{P1 = \:  } \tt{\frac{1}{f}  }

\footnotesize\tt{P1 = \:  } \tt{\frac{1}{0.25}  }

\footnotesize\tt{P1 = \:  } \tt{\frac{100}{25}  }

\footnotesize\tt{P1 = \:  } \tt{ \cancel\frac{100}{25}  }

\footnotesize\tt{P1 = \:  } \tt{ 4 }

\boxed{ \underline{ \underline{\footnotesize\tt{ P1 = \:  } \tt{ 4 D}}}}   \: \red♡

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\footnotesize\tt{2) \: For  \: Concave \:  lens}

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\footnotesize\tt{Focal \:  length(f) =  - 10cm}

\footnotesize\tt{Focal \:  length(f) } \tt{ \: =  \frac{ - 10}{100} m}

\boxed{ \underline{ \underline{\footnotesize\tt{ Focal \:  length(f)= - 0.10m }}}}  \: \red♡

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{P2 = \:  } \tt{\frac{1}{f}  }

\footnotesize\tt{P2 = \:  } \tt{\frac{1}{ - 0.10}  }

\footnotesize\tt{P2 = \:  } \tt{\frac{ - 100}{ 10}  }

\footnotesize\tt{P2 = \:  } \tt{  - 10.0}

\boxed{ \underline{ \underline{\footnotesize\tt{ P2 = \:  } \tt{  - 10 D}}}}  \: \red♡

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{Therefore \:  Power  \: of  \: Combination}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{P =P1 +  P2 }

\footnotesize\tt{ \:  \:   \: =4 +  ( - 10)}

\footnotesize\tt{ \:  \:   \: =4 - 10}

\boxed{ \underline{ \underline{\footnotesize\tt{ \:  \:  \:  = -  6 D}}}} \:  \red♡

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\footnotesize\tt{Now,P = \:  } \tt{ \frac{1}{f}  }

\footnotesize\tt{f=\:  } \tt{ \frac{1}{P}  }

\footnotesize\tt{ f=\:  } \tt{ \frac{1}{ - 6}  \:  m}

\boxed{ \underline{ \underline{\footnotesize\tt{ f=\:  - 0.166m}}}}  \: \red♡

Answered by Ʀɛᴠɛɾíɛ
17

\huge\underline\bold\red{♡Answer♡}

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