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]