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

Question

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

Asked on March 28, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-28T10:22:10.853Z

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: $n_B = 1.6$ $$ \delta_2 = (1.6 - 1)

Accepted Answer · 2026-03-28T10:22:10.853Z

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: $n_B = 1.6$ $$ \delta_2 = (1.6 - 1)