Question

ᴛʜᴇ ᴅɪꜱᴛᴀɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ᴛᴡᴏ ᴄᴏɴꜱᴇᴄᴜᴛɪᴠᴇ ʙʀɪɢʜᴛ ꜰʀɪɴɢᴇꜱ ɪɴ ᴀ ʙᴀᴘᴛɪꜱᴍ ᴇxᴘᴇʀɪᴍᴇɴᴛ ᴜꜱɪɴɢ ʟɪɢʜᴛ ᴏꜰ ᴡᴀᴠᴇʟᴇɴɢᴛʜ 6000 ᴀ⁰ɪꜱ 0.32ᴍᴍ ʙʏ ʜᴏᴡ ᴍᴜᴄʜ ᴡɪʟʟ ᴛʜᴇ ᴅɪꜱᴛᴀɴᴄᴇ ᴄʜᴀɴɢᴇ ɪꜰ ʟɪɢʜᴛ ᴏꜰ ᴡᴀᴠᴇʟᴇɴɢᴛʜ 4800ᴀ⁰ɪꜱ ᴜꜱᴇᴅ?​

Answer 1

$\pmb{ \gray{ \bf{GiVEN}}} :- \\ \rm \tiny{ Distance \: between \: two \: consecutive \: bright \: fringes \: = \: 0.32 \: mm = 0.32 × 10⁻³ m }\\ \rm \: \tiny{ Wavelength \: of \: light \: used = 6000 A⁰ = 6000 × 10⁻¹⁰ m }\\ \\ \\ \pmb{ \gray{ \bf{TO \: \: FiND}}} :- \\ \rm \tiny{Change \: in \: distance \: if \: light \: of \: wavelength \: 4800 A⁰ \: is \: used} \\ \\ \\ \pmb{ \gray{ \bf{SOLUTiON}}} :- \\ \ \tt \: \small{ We \: know \: that} \\ \sf \beta _{i} \propto \lambda_{i}\; and \; \beta_{f}\propto \lambda_{f} \\ \\ \red {\sf \dfrac{0.32\times 10^{-3}}{\beta_{f}} = \dfrac{6000 \times 10^{-10}}{4800 \times 10^{-10}}} \\ \\ \red{\sf \dfrac{0.32\times10^{-3}}{\beta_{f}} = \dfrac{6000}{4800}} \\ \\ \red{\sf\dfrac{0.32\times 10^{-3}}{\beta_{f}} = \dfrac{60}{48} } \\ \\ \red{\sf\dfrac{0.32\times 10^{-3}}{\beta_{f}} = 1.25 } \\ \\ \red{\sf 0.32\times 10^{-3} = 1.25\times\beta_{f}0.32×10 } \\ \\ \red{ \beta_{f} = \sf \dfrac{0.32\times 10^{-3}}{1.25}} \\ \\ \blue{\beta_{f} = \sf 0.256 \times 10^{-3} } \\ \\ \\ \tt \: \therefore \: Change \: \: in \: \: distance :- \\ \\ \scriptsize{\purple{ \sf \beta_f-\beta_i=0.32\times 10^{-3}-0.256\times 10^{-3} = 0.064\times 10^{-3}\: m\:or\:0.064\:mm }}\\ \\ \boxed{ \bf \: Hence \: the \: change \: in \: distance \: is \: 0.064 \: mm.}$

Answer 2

Answer:

Given that,

Wave length of red light λr=6400A∘

Wave length of blue light λb=4800A∘

Fringe width for red light Xr=0.32mm

Fringe width X=dλD

Now, for red light

Xr=dλrD....(I)

For blue light

Xb=dλbD....(II)

Now, dividing equation (II) by(I)

  XrXb=λrλb

 Xb=λrλb×Xr

 Xb=64004800×0.32

 X=0.064mm

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