Optics Question 187
Question: In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1m. The minimum distance between two successive regions of complete darkness is
[IIT JEE (Screening) 2004]
Options:
A) 4 mm
B) 5.6 mm
C) 14 mm
D) 28 mm
Show Answer
Answer:
Correct Answer: D
Solution:
Let nth minima of 400 nm coincides with mth minima of 560 nm then $ (2n-1)400=(2m-1)560 $
Therefore $ \frac{2n-1}{2m-1}=\frac{7}{5}=\frac{14}{10}=\frac{21}{15} $ i.e. 4th minima of 400 nm coincides with 3rd minima of 560 nm.
The location of this minima is $ =\frac{7(1000)(400\times {{10}^{-6}})}{2\times 0.1}=14mm $
Next, 11th minima of 400 nm will coincide with 8th minima of 560 nm Location of this minima is $ =\frac{21(1000)(400\times {{10}^{-6}})}{2\times 0.1}=42mm $ \ Required distance = 28 mm
