Optics Ques 153
- In a Young’s double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $\lambda$. The intensity of light falling on slit $1$ is four times the intensity of light falling on slit $2$ . Choose the correct choice (s).
$(2008,4 M)$
(a) If $d=\lambda$, the screen will contain only one maximum
(b) If $\lambda<d<2 \lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
(c) If the intensity of light falling on slit $1$ is reduced so that it becomes equal to that of slit $2$ , the intensities of the observed dark and bright fringes will increase
(d) If the intensity of light falling on slit $2$ is increased so that it becomes equal to that of slit $1$ , the intensities of the observed dark and bright fringes will increase
Show Answer
Answer:
Correct Answer: 153.(a,b)
Solution:
Formula:
- For $d=\lambda$, there will be only one, central maxima.
For $\lambda<d<2 \lambda$, there will be three maximas on the screen corresponding to path difference, $\Delta x=0$ and $\Delta x= \pm \lambda$.
BETA
