All components of the electromagnetic spectrum in vacuum have the same

151. All components of the electromagnetic spectrum in vacuum have the same

In a vacuum, all components of the electromagnetic spectrum (radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays) travel at the same speed, which is the speed of light, $c$. Their frequencies, wavelengths, and energies, however, differ.

The correct option is (4).
$\boxed{\text{4}}$

152. The focal length of a convex lens of refractive index $\frac{3}{2}$ is 20 cm in air. Focal length of this lens when immersed in a liquid of refractive index of $\frac{5}{2}$ will be

Step 1: Write the lens maker's formula.
The lens maker's formula relates the focal length of a lens to its refractive index, the refractive index of the surrounding medium, and its radii of curvature:

$$\frac{1}{f} = \left(\frac{n_l}{n_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

where $f$ is the focal length, $n_l$ is the refractive index of the lens material, $n_m$ is the refractive index of the surrounding medium, and $R_1$, $R_2$ are the radii of curvature of the lens surfaces.

Step 2: Use the given information for the lens in air to find the term related to radii of curvature.
Given:
Refractive index of lens $n_l = \frac{3}{2}$.
Focal length in air $f_{air} = 20$ cm.
Refractive index of air $n_{air} = 1$.
Substitute these values into the formula:

$$\frac{1}{20} = \left(\frac{3/2}{1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\frac{1}{20} = \left(\frac{3}{2} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\frac{1}{20} = \left(\frac{1}{2}\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Solving for the term $(\frac{1}{R_1} - \frac{1}{R_2})$:

$$\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{2}{20} = \frac{1}{10}$$

Step 3: Calculate the focal length when immersed in the liquid.
Given:
Refractive index of liquid $n_{liquid} = \frac{5}{2}$.
Let the focal length in liquid be $f_{liquid}$.
Substitute $n_l$, $n_{liquid}$, and the calculated $(\frac{1}{R_1} - \frac{1}{R_2})$ into the lens maker's formula:

$$\frac{1}{f_{liquid}} = \left(\frac{3/2}{5/2} - 1\right) \left(\frac{1}{10}\right)$$ $$\frac{1}{f_{liquid}} = \left(\frac{3}{5} - 1\right) \left(\frac{1}{10}\right)$$ $$\frac{1}{f_{liquid}} = \left(\frac{3-5}{5}\right) \left(\frac{1}{10}\right)$$ $$\frac{1}{f_{liquid}} = \left(-\frac{2}{5}\right) \left(\frac{1}{10}\right)$$ $$\frac{1}{f_{liquid}} = -\frac{2}{50} = -\frac{1}{25}$$

Therefore, the focal length in the liquid is:

$$f_{liquid} = -25 \text{ cm}$$

The negative sign indicates that the convex lens behaves as a concave lens when immersed in a medium with a higher refractive index than the lens material.

The correct option is (2).
$\boxed{\text{2}}$

153. Three prisms A, B and C have the prism angle $6^\circ$ but their refractive indices are respectively 1.4, 1.6 and 1.8. If $\delta_1, \delta_2, \delta_3$ be their respective angles of deviation then

Step 1: Recall the formula for the angle of deviation for a small angle prism.
For a prism with a small angle $A$, the angle of deviation $\delta$ is given by:

$$\delta = (n-1)A$$

where $n$ is the refractive index of the prism material.

Step 2: Calculate the angle of deviation for each prism.
Given: Prism angle $A = 6^\circ$ for all three prisms.
For prism A: $n_A = 1.4$

$$\delta_1 = (1.4 - 1) \times 6^\circ = 0.4 \times 6^\circ = 2.4^\circ$$

For prism B: $n_B = 1.6$

$$$$