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Question: The Sun subtends an angle of half a degree at the pole of a concave mirror which has a radius of cur...

The Sun subtends an angle of half a degree at the pole of a concave mirror which has a radius of curvature of 15 m15{\text{ }}m. Then the size (diameter) of the image of the Sun formed by the concave mirror is.
A. 7.5 cm7.5{\text{ }}cm
B. 6.55 cm6.55{\text{ }}cm
C. 3.7 cm3.7{\text{ }}cm
D. 13.1 cm13.1{\text{ }}cm

Explanation

Solution

We have to first draw a diagram of the problem given. By using the concept that when an object is placed at infinity from the surface of a mirror the image forms at the focus of the mirror, we will find the construction of the diagram. Then by using trigonometric concepts we will find the answer.

Complete answer:

The Sun is at infinity which is the object. When an object is at infinity, the image formed due to a concave mirror after reflection from it is at the focus.The pole of the mirror is given as P and the focus as F. Let dd be the diameter of the image of the Sun.In the given, the radius of curvature r=15 mr = 15{\text{ }}m

Let the focal length of the mirror be ff.The relation between radius of curvature rr and focal length ff is f=r2f = \dfrac{r}{2}.
So, the focal length ff of the mirror=152=7.5 m = \dfrac{{15}}{2} = 7.5{\text{ }}m
As per the given question, the sun is subtending 12{\dfrac{1}{2}^ \circ } at the pole.
So, θ2=12(1)\dfrac{\theta }{2} = {\dfrac{1}{2}^ \circ } - - - - - \left( 1 \right)
Converting the given angle in radian we get,
12=π180×12=π360{\dfrac{1}{2}^ \circ } = \dfrac{\pi }{{180}} \times \dfrac{1}{2} = \dfrac{\pi }{{360}} radian(2) - - - - - \left( 2 \right)
Now, from the given triangle AFP we get,
tanθ2=AFFP=d2f\tan \dfrac{\theta }{2} = \dfrac{{AF}}{{FP}} = \dfrac{{\dfrac{d}{2}}}{f}

Now, as tanθ2\tan \dfrac{\theta }{2} is very small, so we consider tanθ2=θ2\tan \dfrac{\theta }{2} = \dfrac{\theta }{2}
Hence, we get,
θ2=d2f\dfrac{\theta }{2} = \dfrac{d}{{2f}}
From(1)\left( 1 \right)we get,
12=d2f{\dfrac{1}{2}^ \circ } = \dfrac{d}{{2f}}
Again, from (2)\left( 2 \right) we get,
π360=d2f\dfrac{\pi }{{360}} = \dfrac{d}{{2f}}
Substituting the value of ff we get,
π360=d2×7.5=d15\dfrac{\pi }{{360}} = \dfrac{d}{{2 \times 7.5}} = \dfrac{d}{{15}}
d=227×24=0.131\therefore d = \dfrac{{22}}{{7 \times 24}} = 0.131
The diameter of the image of the Sun=0.131 m=13.1 cm = 0.131{\text{ }}m = 13.1{\text{ }}cm

Therefore, the correct option is D.

Note: It must be noted that when an object is placed at infinity to a concave mirror the image will always form at the focus of the mirror. The ray of the light is normally considered to be parallel as it comes from infinity. When the angle subtended is very small, then we can omit the trigonometric identity as the difference in value is almost the same.