Question

A star is $100$ times brighter than a star . Then ${m_B} - {m_A}\;$ the difference in their apparent magnitude is-
A. $100$
B. $0.01$
C. $5$
D. $0.2$

Answer 1

We should know that the difference in thin apparent magnitude is equal to the log value of thin intensities.
$\therefore {m_B} - {m_A} = - 2.5\log \left[ {\dfrac{{{I_B}}}{{{I_A}}}} \right]$

Complete step by step answer:
However, the brightness of a star depends on its composition and how far it is from the planet. Astronomers define star brightness in terms of apparent magnitude — how bright the star appears from Earth — and absolute magnitude — how bright the star appears at a standard distance of $32.6$ light-years, or 10 parsecs.
On this magnitude scale, a brightness ratio of $100$is set to correspond exactly to a magnitude difference of 5. As magnitude is a logarithmic scale, one can always transform a brightness ratio $\dfrac{{{I_B}}}{{{I_A}}}$into the equivalent magnitude difference ${m_2} - {m_1}\;$by the formula:
$\therefore {m_B} - {m_A} = - 2.5\log \left[ {\dfrac{{{I_B}}}{{{I_A}}}} \right]$

We know that the difference in their apparent magnetite is equal to the log value of their intensities. Thus,
${m_B} - {m_A} = - 2.5\log \left[ {\dfrac{{{I_B}}}{{{I_A}}}} \right]$- - - - - - - - - - - - - - - - - - - - (1)

${m_B} - {m_A} = - 2.5 \times \log \left( {\dfrac{{1 \times {I_B}}}{{100 \times {I_B}}}} \right)..........(2) \\\ \Rightarrow \left[ {{I_A} = 100\;{\text{as it is 100 times brighter}}{\text{.}}} \right] \\\$
${I_B}$ = intensity of B
$I_A$ = intensity of A

${m_B} - {m_A} = 2.5 \times \left[ {\log 1 - \log 100} \right] \\\ or\;{{\text{m}}_{\text{B}}}{\text{ - }}{{\text{m}}_{\text{A}}}{\text{ = - 2}}{\text{.5}} \times \left( { - 2} \right) \\\ \left[ {{m_B} - {m_A} = 5} \right] \\\$- - - - - - - - - - - - - - - - - - - - (3)

So, the correct answer is “Option C”.

Note:
We should know that the magnitude in astronomy measures the brightness of a star or other celestial body. The brighter the object, the number assigned as a magnitude.