Monochromatic light falls on a grating with grating element 1.0 mm at an glancing angle 1o. When it is diffracting at an glancing angle 3.0o the maximum of first order occurs the λ is a. 0.5µm b. 34µm c. 14µm d. 0.8µm?

Question

Answer

Answer:

Given:
- Grating element, d = 1.0 mm
- Glancing angle of incidence, θ1 = 1°
- Glancing angle of diffraction, θ2 = 3°
- Maximum of first order occurs at λ

To find:
The wavelength of the incident monochromatic light

Solution:

Step 1: Calculate the distance between two consecutive slits in the grating using the formula:
d sin θ = mλ
where,
m = order of diffraction
λ = wavelength of light
d = grating element
θ = angle of diffraction

For maximum of first order, m = 1
So, d sin θ = λ
=> d = λ/sin θ
=> 1.0 × 10^-3 m = λ/sin 3°

Step 2: Convert the grating element to meters for convenience
d = 1.0 mm = 1.0 × 10^-3 m

Step 3: Substitute the value of d in the equation obtained in step 1
1.0 × 10^-3 m = λ/sin 3°
=> λ = 1.0 × 10^-3 m × sin 3°
=> λ ≈ 0.052 µm

Step 4: Compare the calculated value of λ with the given options
The calculated value of λ ≈ 0.052 µm, which is closest to option (a) 0.5 µm.

Therefore, the correct answer is option (a) 0.5 µm.

Answer

Find the answer

Similar IIT JAM Doubts

Which of the following statements holds in (0, 2) if the function y = In (3x4 2x3 6x2 + 6x + 1)

The maximum number of intensity minimum that can be observed in the Fraunhofer diffraction pattern of a single slit (width = 10 µm) illuminated by a laser beam (wavelength 0.630 µm) will be a. 4 b. 7 c. 12 d. 15?

In a grating with grating constant d =a + b, where a is the slit width and b is the separation between the slits, the diffraction pattern has the fourth order missing. The value of b/ais ________.(Specify your answer as an integer.) Correct answer is between '3,3'. Can you explain this answer?

Light of wavelength 680 nm is incident normally on a diffraction grating having 4000 lines/cm. The diffraction angle (in degrees) corresponding to the third-order maximum is ________.(Round off to 2 decimal places) Correct answer is between '54.67,54.71'. Can you explain this answer?

Threshold wavelength for photoelectric emission from a material is 5200 angstrom. Photoelectrons will be emitted when the material is illuminated with monochromatic radiation from:(

Top Courses for IIT JAM

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams