JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 12)
In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle $${1 \over {40}}$$ 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 -
400 nm, 500 nm
625 nm, 500 nm
380 nm, 500 nm
380 nm, 525 nm
Explanation
Path difference = d sin$$\theta $$ $$ \approx $$ d$$\theta $$
= 0.1 $$ \times $$ $${1 \over {40}}$$ mm = 2500nm
or bright fringe, path difference must be integral multiple of $$\lambda $$.
$$ \therefore $$ 2500 = n$$\lambda $$1 = m$$\lambda $$2
$$ \therefore $$ $$\lambda $$1 = 625 (from n = 4), $$\lambda $$2 = 500 (from m = 5)
= 0.1 $$ \times $$ $${1 \over {40}}$$ mm = 2500nm
or bright fringe, path difference must be integral multiple of $$\lambda $$.
$$ \therefore $$ 2500 = n$$\lambda $$1 = m$$\lambda $$2
$$ \therefore $$ $$\lambda $$1 = 625 (from n = 4), $$\lambda $$2 = 500 (from m = 5)
Comments (0)
