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vagabundo [1.1K]
3 years ago
13

A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is t

he magnification
Physics
1 answer:
GuDViN [60]3 years ago
6 0

Answer:

The magnification is  m =  12

Explanation:

From the question  we are told that

   The object distance is u  = 36.2 \ cm

     The focal length is  v  =  39.5 \ cm

From the lens equation we have that

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

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

substituting values

       \frac{1}{v}  =  \frac{1}{39.5}  - \frac{1}{36.2}

       \frac{1}{v}  =  -0.0023

=>   v =  \frac{1}{0.0023}

=>   v =-433.3 \ cm

The magnification is mathematically represented as

         m =-   \frac{v}{u}

substituting values

        m =-   \frac{-433.3}{36.2}

         m =  12

         

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