With respect to air critical angle in a medium for light of red colour `[lambda_(1)]` is `theta`. Other facts remaining same, critical angle for light of yellow colour `[lambda_(2)]` will be

Options

A. `theta`
B. More than `theta`
C. Less than `theta`
D. `(thetalambda_(1))/lambda_(2)`

Answer

To solve the problem regarding the critical angle for light of different colors (red and yellow), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Critical Angle**:
The critical angle (Ic) is defined as the angle of incidence above which total internal reflection occurs. It is related to the refractive index (μ) of the medium by the formula:
\[ \sin(I_c) = \frac{1}{\mu} \]

2. **Given Information**:
- The critical angle for red light (wavelength λ1) is given as θ.
- We need to find the critical angle for yellow light (wavelength λ2).

3. **Wavelength Comparison**:
It is known that red light has a longer wavelength than yellow light:
\[ \lambda_1 > \lambda_2 \]

4. **Refractive Index Relation**:
The refractive index (μ) of a medium is inversely proportional to the wavelength of light:
\[ \mu \propto \frac{1}{\lambda} \]
Therefore, since λ1 > λ2, we can conclude:
\[ \mu_1 < \mu_2 \]
where μ1 is the refractive index for red light and μ2 is the refractive index for yellow light.

5. **Critical Angle for Yellow Light**:
Since μ2 > μ1, we can analyze the critical angles:
\[ \sin(I_{C1}) = \frac{1}{\mu_1} \quad \text{(for red light)} \]
\[ \sin(I_{C2}) = \frac{1}{\mu_2} \quad \text{(for yellow light)} \]
Because μ2 > μ1, it follows that:
\[ \frac{1}{\mu_2} < \frac{1}{\mu_1} \]
Thus:
\[ \sin(I_{C2}) < \sin(I_{C1}) \]

6. **Conclusion**:
Since the sine of the critical angle for yellow light is less than that for red light, we can conclude that:
\[ I_{C2} < I_{C1} \]
Therefore, the critical angle for yellow light (I_{C2}) is less than θ (the critical angle for red light).

### Final Answer:
The critical angle for yellow light will be **less than θ**.