JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 23)
A prism of refractive index n1 and another prism of refractive index n2 are stuck together (as shown in the figure). n1 and n2 depend on $$\lambda$$, the wavelength of light, according to the relation
$${n_1} = 1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$$ and $${n_2} = 1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$$
The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be _____________ nm.
_27th_July_Morning_Shift_en_23_1.png)
$${n_1} = 1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$$ and $${n_2} = 1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$$
The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be _____________ nm.
Answer
600
Explanation
For no bending, n1 = n2
$$1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$$ = $$1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$$
On solving,
9 $$\times$$ 10$$-$$14 = 25$$\lambda$$2
$$\lambda$$ = 6 $$\times$$ 10$$-$$7
$$\lambda$$ = 600 nm
$$1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$$ = $$1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$$
On solving,
9 $$\times$$ 10$$-$$14 = 25$$\lambda$$2
$$\lambda$$ = 6 $$\times$$ 10$$-$$7
$$\lambda$$ = 600 nm
Comments (0)
