Wavelength of red light is 7-5 - 10-5 cm and that of blue light is 5-0 - 10-5 cm- The value of n for which n-1 th blue bright band coincides with nth red band- is-

The correct option is D 2 Position of bright Band in YDSE. y=nDλd So, By given problem n Dλ _Rd=(n+1)Dλ_ Bd n7.5× 10^ 5=(n+1)5× 10^ 5 7. ...

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Standard XII

Physics

Intensity Explanation in YDSE

Wavelength of...

Question

Wavelength of red light is $7.5 \times 10^{- 5} c m$ and that of blue light is $5.0 \times 10^{- 5} c m$ . The value of $n$ for which $(n + 1)$ th blue bright band coincides with $n^{t h}$ red band, is:

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Solution

The correct option is D $2$
Position of bright Band in YDSE.
$y = \frac{n D λ}{d}$
So, By given problem
$\frac{n D λ_{R}}{d} = (n + 1) \frac{D λ_{B}}{d}$
$n 7.5 \times 10^{- 5} = (n + 1) 5 \times 10^{- 5}$
$7.5 n = 5 n + 5$
$2.5 n = 5$
$n = 2$

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Wavelength of red light is 7.5×10−5 cm and that of blue light is 5.0×10−5 cm. The value of n for which (n+1) th blue bright band coincides with nth red band, is:

The correct option is D 2Position of bright Band in YDSE.y=nDλdSo, By given problemnDλRd=(n+1)DλBdn7.5×10−5=(n+1)5×10−57.5n=5n+52.5n=5n=2

Wavelength of red light is $7.5 \times 10^{- 5} c m$ and that of blue light is $5.0 \times 10^{- 5} c m$ . The value of $n$ for which $(n + 1)$ th blue bright band coincides with $n^{t h}$ red band, is:

The correct option is D $2$
Position of bright Band in YDSE.
$y = \frac{n D λ}{d}$
So, By given problem
$\frac{n D λ_{R}}{d} = (n + 1) \frac{D λ_{B}}{d}$
$n 7.5 \times 10^{- 5} = (n + 1) 5 \times 10^{- 5}$
$7.5 n = 5 n + 5$
$2.5 n = 5$
$n = 2$