Two stars emit maximum radiation of wavelength 3600 A- and 4800 A- respectively- The ratio of their temperatures is-

Star $S_{1}$ emits maximum radiation of wavelength $420 n m$ and the star $S_{2}$ emits maximum radiation of wavelength $560 n m$ . What is the ratio of the temperature of $S_{1}$ and $S_{2}$ ?

Q. Two stars A and B radiate maximum energy at the wavelengths of

360

nm and

480

nm respectively. Then the ratio of the surface temperatures of A and B is.

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Two stars emit maximum radiation of wavelength 3600 A∘ and 4800 A∘ respectively. The ratio of their temperatures is-

The correct option is D 4:3By Wein′s displacement lawAt maximum radiation λT=constant⇒λ1T1=λ2T2⇒T1T2=λ2λ1Ratio of temperatures=T1T2=43

Two stars emit maximum radiation of wavelength $3600$ $A^{\circ}$ and $4800$ $A^{\circ}$ respectively. The ratio of their temperatures is-

The correct option is D 4:3
By $W e i n^{'} s$ $d i s p l a c e m e n t$ law
At maximum radiation $λ T = c o n s t a n t$
$\Rightarrow λ_{1} T_{1} = λ_{2} T_{2}$
$\Rightarrow \frac{T_{1}}{T_{2}} = \frac{λ_{2}}{λ_{1}}$
Ratio of temperatures $= \frac{T_{1}}{T_{2}} = \frac{4}{3}$