To understand the given assertion and reason, we need to evaluate both the behavior of light in an optically denser medium and the principles of Young's double slit experiment.
Assertion (A): When Young's double slit experiment is conducted in an optically denser medium than air, the consecutive fringes come closer.
Reason (R): The speed of light reduces in an optically denser medium than air while its frequency does not change.
Analysis:
1. The fringe separation (or fringe width, w) in Young's double slit experiment is given by the formula:
w = (λD) / d
Where λ is the wavelength of light, D is the distance between the slits and the screen, and d is the distance between the slits.
2. In an optically denser medium, the speed of light v is less than in air, while the frequency f of the light remains constant. Therefore, the wavelength in the medium λ' is given by:
λ' = v / f
3. The wavelength in the denser medium λ' is related to the wavelength in air λ by the refractive index n of the medium:
λ' = λ / n
4. Substituting λ' in the fringe width formula:
w' = (λ' D) / d = (λD) / (nd)
5. Since n > 1 in an optically denser medium, w' < w. This means the fringes are closer together in a denser medium compared to air.
Conclusion: Both the assertion (A) and the reason (R) are true. The density of the medium affects the speed of light, and hence the wavelength, which in turn affects the fringe spacing. The reduction in fringe spacing, as stated in the assertion, is correctly explained by the reason.
Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Concept:
In Young’s Double Slit Experiment (YDSE), the fringe width is given by:
$$ \beta = \dfrac{\lambda D}{d} $$ where
$\lambda$ = wavelength of light in the medium,
$D$ = distance between slits and screen,
$d$ = distance between the two slits.
In a denser medium, the wavelength decreases because:
$$ \lambda = \dfrac{v}{f} $$ and since the speed of light (v) decreases while frequency (f) remains constant, $\lambda$ becomes smaller.
Hence, the fringe width $\beta$ also decreases, which means the fringes come closer together.
Assertion (A): True — In an optically denser medium, consecutive fringes come closer.
Reason (R): True — Speed of light decreases, frequency remains the same.
Moreover, (R) correctly explains (A).
✅ Correct Answer: Option 1 — Both (A) and (R) are true and (R) is the correct explanation of (A)
A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is: 
The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)
Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below: