Discussion
Consider two slits are made 3 mm apart and a screen is placed 1.5 m away. When the blue-green light of wavelength 500 nm is incident on the slits, the fringe width observed on the screen is:
Answer: C
The fringe width in Young's experiment is \(\beta = \lambda D / d\).
\[ \begin{aligned} \beta \\ &= \frac{500 \times 10^{-9} \times 1.5}{3 \times 10^{-3}} \\ &= 2.5 \times 10^{-4}\ \text{m} = 0.25\ \text{mm} \end{aligned} \]
The slit separation of 3 mm must be converted to metres and the wavelength of 500 nm to \(5 \times 10^{-7}\) m. The values 0.25 cm and 2.5 cm are ten and a hundred times too large, and 0.25 m is a thousand times too large, which would come from mixing up millimetres and metres. The answer is 0.25 mm, option C.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development