
What is the luminous intensity of the sun if it produces the same illuminance on the earth as produced by a bulb of $10000cd$ at a distance of $0.3m$. The distance between the sun and the earth is $1.5 \times {10^{11}}m$.
(A) $2.5 \times {10^{23}}cd$
(B) $2.5 \times {10^{19}}cd$
(C) $2.5 \times {10^{27}}cd$
(D) $2.5 \times {10^{36}}cd$
Answer
140.7k+ views
Hint: We are given here with the luminous intensity of the bulb at a distance from the point and we are asked to find the luminous intensity of the sun at other distance but producing the same illuminance on the point as that of the bulb. Thus, we will use the formula for illuminance for both cases and then equate them.
Formulae Used
$\varepsilon = \dfrac{P}{{4\pi {d^2}}}$
Where, $\varepsilon $ is the illuminance, $P$ is the luminous intensity of the object and $d$ is the distance of the object from the illuminated point.
Step By Step Solution
Firstly,
For the bulb,
${\varepsilon _{Bulb}} = \dfrac{{{P_{Bulb}}}}{{4\pi {d_1}^2}}$
And, for the sun,
${\varepsilon _{Sun}} = \dfrac{{{P_{Sun}}}}{{4\pi {d_2}^2}}$
Now,
According to the question, we should equate ${\varepsilon _{Sun}} = {\varepsilon _{Bulb}}$
Thus, we get
$\dfrac{{{P_{Sun}}}}{{{d_2}^2}} = \dfrac{{{P_{Bulb}}}}{{{d_1}^2}}$
Now,
The given values are
${P_{Bulb}} = {10^4}cd$
${d_1} = 3 \times {10^{ - 1}}m$
${d_2} = 1.5 \times {10^{11}}m$
Putting in these values, we get
${P_{Sun}} = 2.5 \times {10^{27}}cd$
Hence, the answer is (C).
Additional Information
Luminous intensity is defined as a quantity which is used for characterizing a light source. It is further defined as the luminous flux per unit solid angle.
The S.I. Unit for luminous intensity is \[candela = lumen{\text{ }}per{\text{ }}steradian\]symbolized as $cd$.
Note: We equated the illuminance in both the cases as according to the question both the illumination due to the sun and bulb were the same. Further, we cancelled the common terms and finally came up with a relation in which all of the given parameters got connected with each other and thus we can solve the question with ease.
Formulae Used
$\varepsilon = \dfrac{P}{{4\pi {d^2}}}$
Where, $\varepsilon $ is the illuminance, $P$ is the luminous intensity of the object and $d$ is the distance of the object from the illuminated point.
Step By Step Solution
Firstly,
For the bulb,
${\varepsilon _{Bulb}} = \dfrac{{{P_{Bulb}}}}{{4\pi {d_1}^2}}$
And, for the sun,
${\varepsilon _{Sun}} = \dfrac{{{P_{Sun}}}}{{4\pi {d_2}^2}}$
Now,
According to the question, we should equate ${\varepsilon _{Sun}} = {\varepsilon _{Bulb}}$
Thus, we get
$\dfrac{{{P_{Sun}}}}{{{d_2}^2}} = \dfrac{{{P_{Bulb}}}}{{{d_1}^2}}$
Now,
The given values are
${P_{Bulb}} = {10^4}cd$
${d_1} = 3 \times {10^{ - 1}}m$
${d_2} = 1.5 \times {10^{11}}m$
Putting in these values, we get
${P_{Sun}} = 2.5 \times {10^{27}}cd$
Hence, the answer is (C).
Additional Information
Luminous intensity is defined as a quantity which is used for characterizing a light source. It is further defined as the luminous flux per unit solid angle.
The S.I. Unit for luminous intensity is \[candela = lumen{\text{ }}per{\text{ }}steradian\]symbolized as $cd$.
Note: We equated the illuminance in both the cases as according to the question both the illumination due to the sun and bulb were the same. Further, we cancelled the common terms and finally came up with a relation in which all of the given parameters got connected with each other and thus we can solve the question with ease.
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