The wave length of light in two liquids x and y is 3500 A° and 7000 A° respectively, then the critical angle of x relative to y will be
Understand the Problem
The question is asking for the critical angle of light passing from one liquid to another, given the wavelengths in each liquid. To solve this, we will use the formula for critical angle based on the refractive indices derived from the wavelengths of light in two different media.
Answer
$30^\circ$
Answer for screen readers
The critical angle of $x$ relative to $y$ is $30^\circ$.
Steps to Solve

Determine the refractive indices The refractive index $n$ is related to the wavelength of light in a medium. It can be expressed as: $$ n = \frac{\lambda_0}{\lambda} $$ where $\lambda_0$ is the wavelength of light in vacuum (which we'll assume as a constant) and $\lambda$ is the wavelength in the medium. Let's denote:
 $\lambda_x = 3500 , \text{Å}$ (wavelength in liquid $x$)
 $\lambda_y = 7000 , \text{Å}$ (wavelength in liquid $y$)
We will calculate the ratio of these wavelengths to get the refractive indices.

Calculate the refractive indices for each liquid We have: $$ n_x = \frac{7000}{3500} = 2 $$ $$ n_y = 1 , (\text{assuming} , y , \text{is the reference medium}) $$

Apply Snell's Law for critical angle The critical angle $\theta_c$ can be found using the formula: $$ \sin(\theta_c) = \frac{n_y}{n_x} $$ Substituting the values: $$ \sin(\theta_c) = \frac{1}{2} $$

Calculate the critical angle To find $\theta_c$, take the inverse sine: $$ \theta_c = \sin^{1}\left(\frac{1}{2}\right) = 30^\circ $$
The critical angle of $x$ relative to $y$ is $30^\circ$.
More Information
The critical angle is the angle of incidence beyond which light cannot pass through the boundary between two mediums and instead reflects back into the denser medium. In this case, light going from liquid $x$ to liquid $y$ reaches a critical angle of $30^\circ$, indicating that any angle greater than this will result in total internal reflection.
Tips
 Confusing the order of liquids when calculating indices, which can lead to incorrect values for $n_x$ and $n_y$.
 Forgetting to convert wavelengths to the same units before calculating indices, which can affect final results.