You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warbler subtend on your retina?

Respuesta :

Answer:

The angle it subtend on the retina is  [tex]\theta_z = 0.44586^o[/tex]    

Explanation:

From the question we are told that

     The length of the warbler is  [tex]L = 14cm = \frac{14}{100} = 0.14m[/tex]

      The distance from the binoculars is    [tex]d = 18cm = \frac{18}{100} = 0.18m[/tex]

        The magnification of the binoculars is  [tex]M =8[/tex]

Without the 8 X binoculars the  angle made with the angular size of the object  is mathematically represented as

          [tex]\theta = \frac{L}{d}[/tex]

        [tex]\theta = \frac{0.14}{0.18}[/tex]

           [tex]= 0.007778 rad[/tex]

Now magnification can be represented mathematically as

         [tex]M = \frac{\theta _z}{\theta}[/tex]

Where [tex]\theta_z[/tex] is the angle the image of the warbler subtend on your retina when the   binoculars i.e the  binoculars zoom.

So

      [tex]\theta_z = M * \theta[/tex]

=>    [tex]\theta_z =8 * 0.007778[/tex]

            [tex]= 0.0622222224[/tex]

Generally the conversion to degrees can be mathematically evaluated as

             [tex]\theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )[/tex]

              [tex]\theta_z = 0.44586^o[/tex]  

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