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

When a light ray passes through a triangular prism, the angle of deviation $\delta$ is determined by a complex, nonlinear relationship with the angle of incidence $i$. As the angle of incidence increases from a small value, the angle of deviation initially decreases, reaching a single, well-defined minimum value where the light ray passes symmetrically through the prism. Once the angle of incidence increases beyond this point, the angle of deviation begins to rise again, forming a continuous, U-shaped curve that reflects the existence of exactly one angle of incidence for which the deviation is at its minimum.

The focal length of a plano-convex lens is determined by its refractive index and the radius of curvature of its spherical surface. We begin by finding the refractive index $\mu$ of the lens material, which is the ratio of the speed of light in a vacuum to the speed in the lens, yielding $\mu = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$. The radius of curvature $R$ of the convex surface relates to the lens dimensions; with a diameter of 6 cm, the aperture radius is 3 cm, and with a central thickness of 0.3 cm, the geometric approximation $R \approx \frac{a^2}{2t}$ gives $R = \frac{3^2}{2 \times 0.3} = 15 \text{ cm}$.

Utilizing the lens maker's equation $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$, we substitute our known values where the curved surface has a radius of $15 \text{ cm}$ and the flat surface has an infinite radius. This simplifies to $\frac{1}{f} = (1.5 - 1)(\frac{1}{15} - 0)$, which calculates as $\frac{1}{f} = \frac{0.5}{15}$. Consequently, the focal length is $30 \text{ cm}$.

Introducing a transparent glass plate into the optical path causes a lateral shift in the image position, effectively pulling the image closer to the lens. To maintain a sharp focus on the film, the object must be repositioned to compensate for this displacement.

The initial configuration uses the lens equation to find the focal length, where the object is at $u = -240$ cm and the image is at $v = 12$ cm:
$\frac{1}{f} = \frac{1}{12} + \frac{1}{240} = \frac{21}{240} \text{ cm}^{-1}$
The glass plate, having a thickness of 1 cm and a refractive index of 1.50, induces a shift calculated by:
$s = t\left(1 - \frac{1}{\mu}\right) = 1 \times \left(1 - \frac{2}{3}\right) = \frac{1}{3} \text{ cm}$

This shift reduces the effective distance of the image from the lens to $v' = 12 - \frac{1}{3} = \frac{35}{3}$ cm. Applying the lens equation again to find the new object distance $u'$ yields:
$\frac{1}{u'} = \frac{1}{v'} - \frac{1}{f} = \frac{3}{35} - \frac{21}{240} = -\frac{1}{560} \text{ cm}^{-1}$
This indicates that the object should be placed 560 cm, or 5.6 m, from the lens.

The relationship between the speed of an image and the speed of an object in a spherical mirror is found by differentiating the mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, with respect to time. This process yields the expression for the velocity of the image, $\frac{dv}{dt} = -(\frac{v}{u})^2 \frac{du}{dt}$, which simplifies to the derivative relation $\frac{dv}{dt} = -(\frac{f}{u-f})^2 \frac{du}{dt}$ when substituting for the image distance in terms of the focal length and object distance.

For this convex mirror, the focal length is 0.2 m, the initial object distance is -2.8 m, and the speed of the object is -15 m/s. Substituting these values into the derived formula gives $\frac{dv}{dt} = -(\frac{0.2}{-2.8 - 0.2})^2 (-15)$. Simplifying this calculation leads to a final result of 1/15 m/s.

The angle of incidence for the light ray is determined by the angle between the ray vector $\vec{A} = 6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}-10\hat{k}$ and the surface normal, which is the unit vector $\hat{k}$ for the $x-y$ plane. First, calculate the magnitude of the ray vector, $|\vec{A}| = \sqrt{(6\sqrt{3})^2 + (8\sqrt{3})^2 + (-10)^2} = \sqrt{108 + 192 + 100} = 20$. The cosine of the angle of incidence $i$ is found using the dot product, $\cos i = \frac{|\vec{A} \cdot \hat{k}|}{|\vec{A}|} = \frac{10}{20} = 0.5$, which corresponds to an incidence angle of $i = 60^{\circ}$.

To determine the angle of refraction $r$, apply Snell's Law, which relates the refractive indices of the media to the angles of incidence and refraction through the equation
$n_1 \sin i = n_2 \sin r$
Using the provided refractive indices $n_1 = 2\sqrt{2}$ and $n_2 = 3\sqrt{3}$ alongside the calculated angle of incidence $i = 60^{\circ}$, the equation is solved for $r$. Substituting these values and performing the calculation leads to an angle of refraction of $r = 45^{\circ}$.

Since the intensity $I$ of the beam is highest at the center and decreases moving outward, the refractive index defined by $\mu(I) = \mu_0 + \mu_2 I$ is greatest along the axis and gradually decreases toward the periphery. This creates a graded optical medium where the region of higher refractive index at the core acts to slow down the wavefronts relative to the outer edges. Consequently, the light rays are continuously refracted toward this denser axial region, causing the initially parallel beam to converge.

The speed of light in a medium is determined by the formula $c = \frac{c_0}{\mu}$, where $\mu$ is the refractive index and $c_0$ is the speed of light in a vacuum. Given the relationship $\mu = \mu_0 + \mu_2 I$, the refractive index is directly dependent on the intensity of the light beam. Since the intensity is greatest at the center and decreases as the radius increases, the refractive index is also at its maximum value along the axis, causing the speed of light $c$ to be at its minimum at that central position.

The relationship between the object distance and the image distance for a convex lens is governed by the lens formula, expressed as $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. A straight line passing through the origin at an angle of $45^{\circ}$ signifies that the x-coordinate and y-coordinate are equal, which in this experimental context implies that the magnitude of the image distance equals the magnitude of the object distance. By substituting the sign convention for a real image formed by a convex lens, where the object distance is $-u$ and the image distance is $v$, we set these magnitudes equal by letting $v = u = 2f$. Substituting these values into the lens formula gives $\frac{1}{2f} - \frac{1}{-2f} = \frac{1}{2f} + \frac{1}{2f} = \frac{2}{2f} = \frac{1}{f}$, which satisfies the equation perfectly. Consequently, the intersection point on the graph, where the experimental curve meets the line, is $(2f, 2f)$.

To achieve the condition where a light ray grazes the wall of a cylindrical rod, the ray must strike the internal side wall at the critical angle. Applying Snell's law at the rod's internal interface with air, where the refractive index is $n = \frac{2}{\sqrt{3}}$, requires that the sine of the critical angle $C$ satisfy $\sin C = \frac{1}{n}$. This leads to $\sin C = \frac{\sqrt{3}}{2}$, which corresponds to a critical angle of $C = 60^{\circ}$. Given the path of the ray from the center of the rod's end face to the side wall, the internal angle of refraction $r$ is complementary to this critical angle, resulting in $r = 90^{\circ} - 60^{\circ} = 30^{\circ}$. Snell's law at the entry face of the rod relates the external incident angle $\theta$ to the refractive index and the internal angle of refraction through the equation $\sin \theta = n \sin r$. Substituting the known values gives $\sin \theta = \frac{2}{\sqrt{3}} \times \sin 30^{\circ}$, and because $\sin 30^{\circ} = \frac{1}{2}$, this calculation simplifies to $\sin \theta = \frac{1}{\sqrt{3}}$, leading to an incident angle of $\theta = \sin^{-1} \left( \frac{1}{\sqrt{3}} \right)$.

The optical behavior of a convex lens is governed by the relation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, where $f$ represents the constant focal length. Plotting this relationship results in a rectangular hyperbola because the variables $v$ and $u$ are inversely coupled through the constant $f$. This curve demonstrates the continuous transition of image position relative to object placement, tracking the specific hyperbolic trajectory characteristic of the lens equation.

When multiple thin lenses are placed in contact, the total power of the optical system is simply the algebraic sum of the individual lens powers, represented by $P = P_1 + P_2$. By adding the power of the first lens, -15 D, to the power of the second lens, +5 D, we find the equivalent power of the combination to be -10 D. Since power is the reciprocal of focal length in meters, defined by $f = \frac{1}{P}$, we calculate the focal length by substituting the combined power of -10 D to get -0.1 meters. Converting this value to centimeters by multiplying by 100 gives a final focal length of -10 cm.

The angle of minimum deviation for a prism is directly related to the refractive index of the material, where a higher refractive index causes a greater degree of bending. Since the refractive index of the glass is 1.520 for red light and 1.525 for blue light, the red light experiences a smaller refractive index than the blue light. Therefore, the angle of minimum deviation for red light, $D_1$, is less than the angle of minimum deviation for blue light, $D_2$.

The optical power of a thin lens is determined by the lens maker's formula, which depends on the refractive indices of the lens material and the surrounding medium. In air, where the refractive index is 1, the power is expressed as $P_a = \left(\frac{n_g}{1} - 1\right)K$, where $n_g$ is the refractive index of the glass and $K$ is the constant geometric factor $\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Given $n_g = 1.5$ and an air power $P_a = -5 \text{ D}$, we can calculate the geometric factor as $K = \frac{-5}{1.5 - 1} = -10 \text{ m}^{-1}$.

When the lens is placed in a liquid with a refractive index of 1.6, the reciprocal of the focal length in that medium is found using the relation $\frac{1}{f_m} = \left(\frac{1.5}{1.6} - 1\right)K$. Substituting the value of $K$, we get $\frac{1}{f_m} = \left(\frac{1.5 - 1.6}{1.6}\right)(-10) = \left(\frac{-0.1}{1.6}\right)(-10) = 0.625 \text{ m}^{-1}$. This yields a focal length of $f_m = \frac{1}{0.625} = 1.6 \text{ m}$. Applying the definition of optical power in a medium, $P_m = \frac{n_m}{f_m}$, we find $P_m = \frac{1.6}{1.6} = 1 \text{ D}$.

Light rays entering the water from the air can only reach the fish if their angle of incidence is within the critical angle, $\theta_c$, beyond which rays undergo total internal reflection. This limitation creates a circular window of vision for the fish, defined by the critical angle where $\sin \theta_c = \frac{1}{\mu}$. With a refractive index $\mu = 4/3$, we find $\sin \theta_c = 3/4$, which corresponds to a right triangle with an opposite side of 3 and a hypotenuse of 4, resulting in an adjacent side of $\sqrt{4^2 - 3^2} = \sqrt{7}$. Consequently, the tangent of the critical angle is $\tan \theta_c = \frac{3}{\sqrt{7}}$, and since the fish is at a depth of 12 cm, the radius of the circular horizon is $R = 12 \times \frac{3}{\sqrt{7}}$, which equals $\frac{36}{\sqrt{7}}$ cm.

Total internal reflection occurs when light traveling from a denser medium toward a rarer medium strikes the interface at an angle of incidence that exceeds the critical angle, where the critical angle $i_c$ satisfies $\sin(i_c) = 1/n$. Given that the light ray strikes the glass-air interface at an angle of 45 degrees and undergoes total internal reflection, the incident angle must be greater than the critical angle, which means $\sin(45^\circ) > \sin(i_c)$. Substituting the expression for the critical angle into this inequality leads to $\sin(45^\circ) > 1/n$, or $1/\sqrt{2} > 1/n$. This inequality simplifies to the final condition that $n > \sqrt{2}$.

When a plano-convex lens is silvered on its curved surface, the system functions as a concave mirror because light passes through the lens, reflects off the silvered back surface, and refracts through the lens a second time. The effective focal length $F$ of this combination is determined by the equation $\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m}$, where $f_l$ is the focal length of the lens and $f_m$ is the focal length of the curved silvered surface. Applying the lens maker's formula for a plano-convex lens with a refractive index of 1.5 and a radius of curvature of 30 cm, the inverse of the lens focal length is $\frac{1}{f_l} = (1.5 - 1)(\frac{1}{\infty } - \frac{1}{-30}) = \frac{1}{60}$. The curved silvered surface acts as a concave mirror with a radius of 30 cm, meaning its focal length is 15 cm, or $\frac{1}{f_m} = \frac{1}{15}$. Substituting these values into the combination formula results in $\frac{1}{F} = 2(\frac{1}{60}) + \frac{1}{15} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$, which establishes an effective focal length of 10 cm. A real image of the same size as the object is formed by a concave mirror only when the object is placed at the center of curvature, located at twice the focal length, which is 20 cm.

The number of images formed by two plane mirrors inclined at an angle $\theta$ is governed by the ratio of the full circle to that angle. When the division $\frac{360}{\theta}$ yields an even number, the total image count is calculated as $n = \frac{360}{\theta} - 1$. To obtain three images, we set $n=3$, which results in the equation $\frac{360}{\theta} - 1 = 3$. Solving this simplifies to $\frac{360}{\theta} = 4$, meaning the mirrors must be placed at an angle of 90 degrees.

A compound microscope utilizes a short focal length objective lens to inspect small specimens by placing the object just outside the focal point $f$. This specific optical arrangement causes light rays to converge on the opposite side of the lens, resulting in an image that is real, inverted, and enlarged. This initial magnification is crucial for the overall function of the system, as the resulting image serves as the source for the eyepiece to produce an even larger final view.

Optical fibres rely on the principle of total internal reflection to send light through the core without any loss. When light strikes the interface between the core and the cladding at an angle of incidence greater than the critical angle ($i > i_c$), it reflects entirely back into the fibre instead of escaping. This continuous phenomenon allows light signals to be guided efficiently across long distances.

The resolving power of an optical instrument is inversely proportional to the wavelength of the light used, which implies that shorter wavelengths lead to higher resolving power. Based on this inverse relationship, the ratio of the resolving powers for two wavelengths is given by:

$\frac{(RP)_1}{(RP)_2} = \frac{\lambda_2}{\lambda_1}$

By substituting $\lambda_1$ = 4000 Å and $\lambda_2$ = 5000 Å into this expression, we find the ratio to be $\frac{5000}{4000}$, which simplifies to $\frac{5}{4}$.