Ray Optics And Optical Instruments – JEE Main PhysicsPractice Questions & PYQs

The ability of an astronomical telescope to distinguish between two closely spaced objects depends primarily on its resolving power. This property is governed by the relationship $R \cdot P = \frac{D}{1.22 \lambda}$, where $D$ stands for the diameter of the objective lens or aperture, and $\lambda$ represents the wavelength of the light being collected. Since the resolving power scales directly with the diameter, increasing the aperture size provides higher resolution, allowing the observer to see distinct images rather than a blurred spot. By utilizing a large aperture, the telescope maximizes its capacity to capture fine details in the night sky.

The number of images produced by two plane mirrors inclined at an angle $\theta$ is determined by the ratio of a full circle to that angle, expressed as $\frac{360^\circ}{\theta}$. When the angle between the mirrors is 60 degrees, this ratio is calculated as $\frac{360^\circ}{60^\circ} = 6$. Because this quotient is an even integer, the rule dictates subtracting one to account for the overlapping images, which results in a total of 5.

When light passes through a prism at the angle of minimum deviation, the path of the ray becomes symmetrical, requiring the angle of incidence to be equal to the angle of emergence. For an equilateral prism, the prism angle $A$ is $60^{\circ}$. We determine the minimum deviation $\delta_{\text{min}}$ using the refractive index formula $\mu = \frac{\sin \left( \frac{A+\delta_{\text{min}}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$. Substituting the given values $\mu = \sqrt{2}$ and $A = 60^{\circ}$ leads to the equation $\sqrt{2} = \frac{\sin \left( \frac{60^{\circ}+\delta_{\text{min}}}{2} \right)}{\sin 30^{\circ}}$.

Since $\sin 30^{\circ}$ equals $0.5$, the equation simplifies to $\sin \left( \frac{60^{\circ}+\delta_{\text{min}}}{2} \right) = \frac{\sqrt{2}}{2}$, which is equivalent to $\frac{1}{\sqrt{2}}$. This indicates that the angle $\frac{60^{\circ}+\delta_{\text{min}}}{2}$ must be $45^{\circ}$, solving for a minimum deviation $\delta_{\text{min}}$ of $30^{\circ}$. Applying the relationship for minimum deviation where $e = \frac{A+\delta_{\text{min}}}{2}$, we calculate the emergence angle as $\frac{60^{\circ}+30^{\circ}}{2}$, resulting in $45^{\circ}$.

The speed of light propagating through a medium is determined by the magnetic permeability $\mu$ and electric permittivity $\epsilon$ of that material, expressed as $v = \frac{1}{\sqrt{\mu \epsilon}}$. This relationship can be written in terms of the vacuum speed of light $c$ as $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$, where $\mu_r$ and $\epsilon_r$ represent the relative permeability and relative permittivity of the medium, respectively.

Calculating the velocity $v_2$ for the second medium involves evaluating the ratio $\frac{c}{\sqrt{9}}$. Substituting the vacuum speed $c = 3 \times 10^8 \text{ m/s}$ into this expression results in $v_2 = \frac{3 \times 10^8 \text{ m/s}}{3}$, which simplifies to $1.0 \times 10^8 \text{ m/s}$.

The lateral displacement of a light ray passing through a parallel-sided glass plate is determined by the refractive index of the glass and the angle of refraction within the material. Given that the angle of incidence $i$ is equal to the critical angle, we use the relationship $\sin i = 1/\mu$. With a refractive index $\mu = \sqrt{2}$, this gives $\sin i = 1/\sqrt{2}$, establishing that the incidence angle is $45^{\circ}$. Applying Snell's Law at the first interface, $1 \cdot \sin i = \mu \sin r$, we calculate $\sin 45^{\circ} = \sqrt{2} \sin r$, which simplifies to $1/\sqrt{2} = \sqrt{2} \sin r$, identifying the refraction angle $r$ as $30^{\circ}$.

The lateral displacement $x$ is defined by the formula $x = t \sin(i - r) \sec r$, where $t = \sqrt{3}$ cm represents the thickness of the plate. Substituting the known values leads to $x = \sqrt{3} \sin(45^{\circ} - 30^{\circ}) \sec 30^{\circ}$, which expands to $x = \sqrt{3} \sin 15^{\circ} \frac{1}{\cos 30^{\circ}}$. Given $\sin 15^{\circ} = 0.26$ and knowing that $\cos 30^{\circ} = \sqrt{3}/2$, the expression becomes $x = \sqrt{3} \cdot 0.26 \cdot \frac{2}{\sqrt{3}}$. Simplifying the terms results in $x = 0.52$ cm, or $52 \times 10^{-2}$ cm.

The angle of incidence is determined by analyzing the incident ray vector $\vec{P} = 4\sqrt{3}\hat{i} - 3\sqrt{3}\hat{j} - 5\hat{k}$ and its orientation relative to the normal of the boundary, which is the $z$-axis ($\hat{k}$). The magnitude of this vector is calculated as $|\vec{P}| = \sqrt{(4\sqrt{3})^2 + (-3\sqrt{3})^2 + (-5)^2}$, which simplifies to $\sqrt{48 + 27 + 25} = 10$. Since the normal to the $xy$-plane aligns with the $z$-axis, the cosine of the angle of incidence $i$ is found using the ratio of the $z$-component of the vector to its total magnitude: $\cos i = \frac{|-5|}{10} = \frac{1}{2}$, confirming that $i = 60^{\circ}$.

With the angle of incidence established, Snell's Law, $\mu_1 \sin i = \mu_2 \sin r$, allows us to find the angle of refraction $r$. Utilizing the refractive indices $\mu_1 = \sqrt{2}$ and $\mu_2 = \sqrt{3}$, we substitute these values and the angle of incidence into the equation: $\sqrt{2} \sin 60^{\circ} = \sqrt{3} \sin r$. Simplifying this expression, we get $\sqrt{2} (\frac{\sqrt{3}}{2}) = \sqrt{3} \sin r$, which reduces to $\frac{\sqrt{2}}{2} = \sin r$, or $\sin r = \frac{1}{\sqrt{2}}$. This calculation results in an angle of refraction $r = 45^{\circ}$. The final difference between the angle of incidence and the angle of refraction is $60^{\circ} - 45^{\circ} = 15^{\circ}$.

When a light source is placed at the bottom of a tank, it creates a circular illuminated patch on the water's surface. This phenomenon occurs because light rays hitting the water-air interface at an angle greater than the critical angle undergo total internal reflection and cannot escape the water. The light that does emerge forms a cone with its apex at the bulb, and the radius of the resulting circular area on the surface is determined by the depth $h$ and the critical angle $i_c$ according to the relation $r = h \tan i_c$.

The critical angle $i_c$ is defined by the refractive index $\mu$ such that $\sin i_c = 1/\mu$. With the refractive index of water given as $4/3$, we find $\sin i_c = 3/4$. Using the trigonometric identity $\tan i_c = \frac{\sin i_c}{\sqrt{1 - \sin^2 i_c}}$, we calculate $\tan i_c = \frac{3/4}{\sqrt{1 - 9/16}} = \frac{3/4}{\sqrt{7/16}} = \frac{3}{\sqrt{7}}$. Given the depth of the tank $h = \sqrt{7}$ meters, the radius of the illuminated circle is $r = \sqrt{7} \times \frac{3}{\sqrt{7}} = 3$ meters. Consequently, the area of the surface is $\pi r^2 = \pi (3)^2 = 9\pi$ square meters, which indicates that $x$ is 9.

Snell's law dictates that the product of the refractive index and the sine of the angle relative to the normal remains constant at an interface between two optical media. With light originating in air, where the refractive index is 1, and entering a medium with a refractive index of $\sqrt{2n}$, the relationship between the incident angle $\theta$ and the refracted angle $\frac{\theta}{2}$ is expressed as $1 \cdot \sin(\theta) = \sqrt{2n} \cdot \sin\left(\frac{\theta}{2}\right)$.

Applying the trigonometric double-angle identity $\sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$ allows for the simplification of the equation:
$2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) = \sqrt{2n}\sin\left(\frac{\theta}{2}\right)$
Dividing both sides of this equality by $\sin\left(\frac{\theta}{2}\right)$ yields $2\cos\left(\frac{\theta}{2}\right) = \sqrt{2n}$. Simplifying this result leads to $\cos\left(\frac{\theta}{2}\right) = \frac{\sqrt{2n}}{2}$, which is equivalent to $\sqrt{\frac{2n}{4}}$ or $\sqrt{\frac{n}{2}}$. Consequently, the incident angle is determined as $\theta = 2\cos^{-1}\left(\sqrt{\frac{n}{2}}\right)$.

The path of light through a glass slab is governed by Snell's law, where the refractive index of air is taken as 1. Applying the law gives $1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin r$, which simplifies to $\sin r = \frac{1}{2}$, meaning the angle of refraction is $30^{\circ}$.

This refraction causes a lateral shift between the incident and emergent rays, defined by the formula $d = \frac{t \sin(i - r)}{\cos r}$, where $t$ is the thickness of the slab. Substituting the known lateral shift of $4\sqrt{3} \text{ cm}$ and the angles into the equation results in $4\sqrt{3} = \frac{t \sin(60^{\circ} - 30^{\circ})}{\cos 30^{\circ}}$. Since $\sin 30^{\circ} = \frac{1}{2}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, the expression becomes $4\sqrt{3} = t \cdot \tan 30^{\circ}$. Because $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, the equation is $4\sqrt{3} = \frac{t}{\sqrt{3}}$, leading to a final thickness of $t = 12 \text{ cm}$.

Refraction between two optical media is fundamentally governed by Snell's Law, which relates the refractive indices of the media to the angles of incidence and refraction. When a light ray emerges from a prism face at an angle of 90 degrees, it indicates the light is grazing the surface, allowing us to solve for an unknown refractive index by equating the product of the refractive index and the sine of the angle of incidence in one medium to the product of the refractive index and the sine of the angle of refraction in the second medium.

Applying this principle to the prism, we use the following equation:
$1.5 \sin 60^\circ = \frac{\sqrt{x}}{4} \sin 90^\circ$
Substituting the trigonometric values, where $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\sin 90^\circ = 1$, leads to the equation $1.5 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{x}}{4}$. Multiplying both sides of this expression by 4 results in $3 \sqrt{3} = \sqrt{x}$. Squaring both sides of this equality demonstrates that $x = 27$.

An equilateral prism possesses a prism angle of $A = 60^{\circ}$. Under the condition of minimum deviation, light travels symmetrically through the prism, meaning the angle of refraction is equal to half the prism angle, represented by $r = \frac{A}{2} = 30^{\circ}$. By applying Snell's law at the interface, $\sin i = \mu \sin r$, and substituting the given refractive index $\mu = \sqrt{2}$, we find that $\sin i = \sqrt{2} \sin 30^{\circ} = \frac{1}{\sqrt{2}}$, which indicates that the angle of incidence is $i = 45^{\circ}$. Finally, using the relationship $\delta_{min} = 2i - A$, the angle of minimum deviation is calculated as $2(45^{\circ}) - 60^{\circ} = 30^{\circ}$.