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Question: The ratio of the diameter of the sun to the distance between the earth and the sun is approximately ...

The ratio of the diameter of the sun to the distance between the earth and the sun is approximately 0.009. The approximate diameter of the image of the sun formed by a concave spherical mirror of radius of curvature 0.4m is

A)4.5×10−6mA)4.5\times {{10}^{-6}}m

B)4.0×10−6mB)4.0\times {{10}^{-6}}m

C)3.6×10−3mC)3.6\times {{10}^{-3}}m

D)1.8×10−3mD)1.8\times {{10}^{-3}}m

Explanation

Solution

Making the correct ray diagram of the question is halfway into solving the problem. Knowledge of reflection and concave mirror is necessary for this problem to be solved.

Complete Step by step solution:

Making the correct ray diagram of the question is halfway into solving the problem. Knowledge of reflection and concave mirror is necessary for this problem to be solved.

Here Sun refers to the original object in the question. Point A is the point of intersection of the bottommost tip of the object and the horizon. Point O refers to the center of the concave mirror on the horizon. Point B refers to the intersection of the topmost point of the image with the horizon.θi{{\theta }_{i}}is the angle of incidence on the mirror and θr{{\theta }_{r}} is the angle of reflection.

d is the diameter of the sun and D is the distance from the center of the sun to the center of the mirror. From the questiondD=0.009\dfrac{d}{D}=0.009

Using the laws of reflection, angle of incidence = angle of reflection. Therefore, θi=θr=θ.{{\theta }_{i}}={{\theta }_{r}}=\theta .

We can also consider the ratio dD=tan⁡θi=0.009⇒tan⁡θ=0.009\dfrac{d}{D}=\tan {{\theta }_{i}}=0.009\Rightarrow \tan \theta =0.009

We will consider the sun is at a large distance from the mirror, therefore, the image will form at the focus of the mirror. Hence, distance OB = f, where f is the focal length of the concave mirror.

For a spherical concave mirror, focal length = half of the radius of the curvature of the mirror.

Given radius of curvaturer=0.4m,r=0.4m,therefore,f=r2=0.4m2=0.2m.f=\dfrac{r}{2}=\dfrac{0.4m}{2}=0.2m.

Now considering the diameter of the image as di,{{d}_{i}},then,tan⁡θr=dif⇒tan⁡θ=di0.2\tan {{\theta }_{r}}=\dfrac{{{d}_{i}}}{f}\Rightarrow \tan \theta =\dfrac{{{d}_{i}}}{0.2}

Taking the value of tan⁡θ\tan \theta from above,0.009=di0.2⇒di=0.009×0.20.009=\dfrac{{{d}_{i}}}{0.2}\Rightarrow {{d}_{i}}=0.009\times 0.2

Therefore, the diameter of the image di=0.0018m⇒di=1.8×10−3m.{{d}_{i}}=0.0018m\Rightarrow {{d}_{i}}=1.8\times {{10}^{-3}}m.

Therefore, the correct answer is option (D).

Note:

Keep a thorough check of the decimal points. Probability of making any error in decimal points is very high even though the procedure may be correct.

Knowing the relationship between the focal length and the radius of curvature of a spherical mirror is important.