question_answer

In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle \[\frac{1}{40}\] rad by using light of wavelength\[{{\lambda }_{1}}\]. When the light of wavelength \[{{\lambda}_{2}}\] is used a bright fringe is seen at the same angle in the same set up. Given that \[{{\lambda}_{1}}\] and \[{{\lambda}_{2}}\] are in visible range (380 nm to 740 nm), their values are - [JEE Main 10-Jan-2019 Morning]

A)  400 nm, 500 nm

B)       625 nm, 500 nm

C)  380 nm, 500 nm D)       380 nm, 525 nm Correct Answer: B Solution : [b] \[d\text{ }sin\text{ }\theta =n\lambda ,\]

In question \[\theta =\frac{1}{40}\] which is very small

\[\therefore \text{ }sin\text{ }\theta =\theta =\frac{1}{40}\]

\[d\times \theta =n\lambda ,\]

\[n\,\,=\,\,\frac{d\theta }{\lambda }\,\,=\,\,\frac{0.1\,mm}{40\,\lambda }\]

\[n=\frac{2500}{\lambda }nm\,\]

When \[\lambda =380\]

\[{{n}_{1}}\,=\,\frac{2500}{350}\,\,=\,\,6.578\]

\[for\,\,\lambda =740\,nm\]

\[{{n}_{2}}\,=\,\frac{2500}{740}\,=\,3.378\]

\[\therefore \,\,\,\,n=4,\,\,5,\,\,6\]

\[for\text{ }n=4,\text{ }\lambda =625\text{ }nm\]

\[for\text{ }n=5,\text{ }\lambda =500\text{ }nm\]