Question

Find the maximum value of resolving power of a grating 3 cm wide having 5000 lines per cm, if the wavelength of light used is 589 nm.​

Answer 1

Given

• Here we have a grating 3 cm wide with 5000 lines
• Wavelength = 589

To Find

• Resolving Power

Solution

● d = 1/N

● RP = d/λ

✭ Converting Units :

• 1 Nm = 10 Å
• 589 Nm = 589 × 10
• λ = 5890 Å

━━━━━━━━━━━━━━━━━━

✭ Width of one line :

→ d = 1/N

→ d = 1/5000

→ d = 0.0002 cm

or

→ d = 0.0002/100

→ d = 2 × 10⁻⁶ m

✭ Resolving Power :

→ RP = d/λ

→ RP = (2 × 10⁻⁶)/5890 × 10⁻¹⁰

→ RP = 2/(5890 × 10⁻⁴)

→ RP = 3.39 ≈ 3.4

Answer 2

$\huge{\boxed{\bold{Question}}}$Find the maximum value of resolving power of a grating 3 cm wide having 5000 lines per cm, if the wavelength of light used is 589 nm.

Answer

Resolving power of the diffaction grating is given as 3.4

we have to find

RP=????

Explanation:

As we know that the number is of line is 5000 lines per centimetre

So,

We have widtg is one line given as

D=1/5000 cm

D=2×10^-6m

Now we have

$\rm \implies \: \theta \: = \frac{ \lambda}{d} \\ \\ \bold{now \: we \: have \: resolving \: \: \: power \: given \: as} \\ \\ \tt \longmapsto \: rp = \frac{d}{ \lambda} \\ \\ \longmapsto \tt \: rm \: = \frac{2 \times 10 ^{ - 6} }{5890 \times {10}^{ - 10} } \\ \\ \tt \longmapsto \: \mathfrak{rp = 3.4}$

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Topic : resolving power of diffraction grating

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