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

When a light ray passes through a prism with a very small refracting angle, the minimum deviation is calculated using the relation $\delta_{min} = (\mu - 1)A$. Inserting the given refractive index $\mu = 1.5$ and prism angle $A = 1^{\circ}$ into this expression results in $\delta_{min} = (1.5 - 1) \times 1^{\circ} = 0.5^{\circ}$. Since 0.5 is equivalent to 5/10, the value of N is 5.

The relationship between the image distance $v$ and object distance $u$ for a concave lens is governed by the equation $v = \frac{uf}{u+f}$. Analyzing the behavior at the limits shows that as the object distance approaches infinity, the image distance approaches the focal length, which is established by the condition $u = \infty$ resulting in $v = f$. Furthermore, evaluating the case where the image distance equals the object distance, $v=u$, reduces the formula to $u=0$, indicating that the curve must pass through the origin. These specific geometric constraints are uniquely satisfied by the graph shown in option D.

The light from the particle reflects off the concave mirror at the bottom of the water-filled glass. With a radius of curvature of 40 cm, the mirror possesses a focal length of $f = -20$ cm. Because the mirror operates within a medium of refractive index $\mu = 1.33$, the effective object distance $u$ is adjusted by this index, resulting in $u = -(\frac{4}{3}) \times 5 = -\frac{20}{3}$ cm.

Applying the mirror equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, we have $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-\frac{20}{3}}$. Rearranging this gives $\frac{1}{v} = \frac{3}{20} - \frac{1}{20} = \frac{2}{20}$, so the image is formed at a distance $v = 10$ cm from the mirror. To find the final apparent distance $d$ as viewed from above the water surface, we calculate the effective depth using the total distance of the image from the surface relative to the refractive index, providing $d = \frac{10+5}{1.33} = 11.7$ cm.

The relationship between magnification $m$ and image distance $v$ for a thin lens is defined by the equation $m = 1 - \frac{v}{f}$, which represents a straight line with a slope of $-\frac{1}{f}$. Based on the provided graph, the slope is determined by the ratio of the vertical intercept $c$ to the horizontal intercept $b$, which gives a slope of $-\frac{c}{b}$ due to the line's downward direction. Equating these two expressions, $-\frac{1}{f} = -\frac{c}{b}$, allows us to determine the focal length as $f = \frac{b}{c}$.

When two thin lenses are placed in physical contact, the optical system acts as a single lens whose total power is the sum of the individual powers of the component lenses. For the plano-concave lens with refractive index $\mu_1$ and radius of curvature $R$, the power contribution is defined by $\frac{1-\mu_1}{R}$, while the plano-convex lens with refractive index $\mu_2$ and the same radius $R$ contributes a power of $\frac{\mu_2-1}{R}$. Combining these by adding their reciprocals gives the total power as $\frac{1}{F} = \frac{1-\mu_1}{R} + \frac{\mu_2-1}{R}$, which simplifies algebraically to $\frac{\mu_2-\mu_1}{R}$. Taking the reciprocal of this resulting power identifies the final focal length of the lens combination as $\frac{R}{\mu_2-\mu_1}$.

Concave mirrors produce an upright, magnified image only when the object is placed between the focal point and the mirror, resulting in a virtual image behind the surface. Magnification is the ratio of image distance to object distance, which is 5 in this instance. Because the image is virtual, the relationship between the distances is $v = -5u$.

The mirror equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ relates the focal length of 0.4 m to the object and image distances:
$\frac{1}{0.4} = \frac{1}{-5u} + \frac{1}{u}$
Simplifying the right side of this equation by finding a common denominator results in:
$\frac{1}{0.4} = \frac{4}{5u}$
This equation leads to $5u = 1.6$, meaning the object distance is $u = 0.32$ m.

The relationship between the velocities of an object and its image formed by a lens is dictated by the longitudinal magnification, which is derived from the time derivative of the thin lens equation. To determine the velocity of the image, first locate its initial position by applying the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. Using the focal length $f = 0.3$ m and the object position $u = -20$ m, the calculation $\frac{1}{v} = \frac{1}{0.3} - \frac{1}{20} = \frac{10}{3} - \frac{1}{20} = \frac{197}{60}$ gives an image distance of $v = \frac{60}{197}$ m.

With the positions established, find the transverse magnification $m$ using the ratio $\frac{v}{u}$, which results in $m = \frac{60/197}{-20} = -\frac{3}{197}$. The speed of the image $v_I$ is then proportional to the square of this magnification multiplied by the object speed $v_O$, expressed as $|v_I| = m^2 |v_O|$. Substituting the given object speed of $5$ m/s and the square of the magnification, $m^2 = (\frac{3}{197})^2 \approx 2.32 \times 10^{-4}$, results in a value of approximately $1.16 \times 10^{-3}$ m/s. Because the object moves away from the lens, the resulting image must move towards the lens to satisfy the lens equation, confirming the direction of motion.

The field of view of a plane mirror corresponds to the cone of light reflecting from the mirror edges, which is determined by treating the virtual image $S'$ as an effective point source. The virtual image $S'$ forms at a distance $L$ behind the mirror, and since the man walks at a distance $2L$ in front of the mirror, the total distance between the image $S'$ and the man's path is $L + 2L = 3L$. By applying the property of similar triangles, where the width of the mirror $d$ is at a distance $L$ from the image $S'$, the width of the field of view $W$ at a distance $3L$ is given by the ratio $\frac{W}{3L} = \frac{d}{L}$. Solving for $W$ yields $W = 3d$.

The focal length of a biconvex lens is determined by the lens maker's formula, $\frac{1}{f} = (\mu_{\text{rel}}-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, where $\mu_{\text{rel}}$ represents the refractive index of the lens relative to the surrounding medium. Immersing the entire setup in water alters this relative refractive index, which necessarily changes the focal length of the lens. Since the focal length is no longer the same as it was in air, the image can no longer form at the original fixed position, causing it to disappear from the screen.

When a light ray enters an optical fiber, it refracts according to Snell's Law, $n_1 \sin \theta_1 = n_2 \sin \theta_2$, where the ray travels at an angle $\theta_2$ relative to the fiber axis. By using the refractive index of 1.31 and the incident angle of 40 degrees, we find that $\sin \theta_2 = 0.64 / 1.31$. This internal angle determines the geometry of the zigzag path, where the horizontal distance $x$ between consecutive reflections is given by $x = d / \tan \theta_2$. With a total fiber length of 2 m and a diameter of 20 $\mu$m, calculating the ratio of the total length to this horizontal segment distance $x$ shows that the ray makes approximately 57000 reflections.

When light travels through an optical system, the image formed by the first component acts as the object for the subsequent one. Using the thin lens equation $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ with $f = 20$ cm and $u = -40$ cm, the first image is formed at $v = +40$ cm behind the lens, which places it $20$ cm in front of the mirror given the $60$ cm separation. Utilizing the mirror equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ with $f = -10$ cm and $u = -20$ cm, the second image is calculated to be $20$ cm in front of the mirror. Finally, because the product of the magnifications from both stages results in an overall magnification magnitude of $1$, the final image remains the same size as the original object.

The geometry of light reflection between two inclined mirrors creates a triangle where the incident ray, the reflected ray, and the mirrors themselves dictate the angular relationships. Because the incoming ray is parallel to the second mirror, the angle of incidence at the first mirror is equal to the angle between the mirrors, $\theta$, and the law of reflection ensures this same angle persists throughout the path. Summing the internal angles of the triangle formed by the light path shows that $3\theta = 180^\circ$, which determines that the angle between the mirrors is 60°.

The relationship between the focal length of a thin lens, the object distance, and the image distance is governed by the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. Since the lens initially forms a sharp image at 10 cm when the object is placed at 10 cm, we can determine the focal length by substituting these values into the equation as $\frac{1}{10} - \frac{1}{-10} = \frac{1}{f}$, which results in $\frac{2}{10} = \frac{1}{f}$, confirming that the focal length is 5 cm.

When a glass slab is placed in the path of light, it causes an apparent shift in the object's position due to refraction, which is calculated using the formula $\Delta x = t(1 - \frac{1}{\mu})$. Given the slab thickness of 1.5 cm and a refractive index of 1.5, the shift is $\Delta x = 1.5(1 - \frac{2}{3}) = 0.5$ cm. Because this shift occurs in the direction of the incident light rays, the object effectively appears to be 0.5 cm closer to the lens, changing the new object distance to -9.5 cm.

To find the new image position, we apply the lens formula again with the updated object distance: $\frac{1}{v'} - \frac{1}{-9.5} = \frac{1}{5}$. Rearranging this gives $\frac{1}{v'} = \frac{1}{5} - \frac{2}{19}$, which simplifies to $\frac{1}{v'} = \frac{9}{95}$, resulting in an image distance of approximately 10.55 cm. Comparing this new position to the original image distance of 10 cm, the screen must be shifted away from the lens by 0.55 cm.

In an optical system consisting of a convex lens placed on a plane mirror, a real inverted image forms at the same location as the object when light rays reflect back along their original path, implying they strike the mirror surface normally. This occurs when the object distance matches the effective focal length of the system. Initially, with the object at 18 cm, the effective power of the system is $P_{net} = 2P_L$, which, based on the setup, is $P_{net} = 1/9$.

When a liquid layer is introduced between the lens and the mirror, the system gains power from the liquid, which acts as a plano-concave lens. The new effective focal length becomes 27 cm, corresponding to a new system power $P'_{net} = 2/27$. This total power is expressed as $P'_{net} = 2P_L + 2P_{liquid}$, where the liquid power is $P_{liquid} = (\mu_l - 1)(-1/R_L)$ and the lens radius $R_L$ is 18 cm. Substituting the known values leads to the equation $2/27 = 1/9 + 2(\mu_l - 1)(-1/18)$. Simplifying this expression gives $2/27 - 1/9 = -(\mu_l - 1)/9$, which confirms the liquid refractive index $\mu_l = 4/3$.

Solving a multi-lens system requires treating the image produced by the first lens as the object for the subsequent lens. By applying the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ to the first lens with a focal length of 5 cm and an object at $-20$ cm, the equation $\frac{1}{v_1} - \frac{1}{-20} = \frac{1}{5}$ reveals that $\frac{1}{v_1} = \frac{3}{20}$, placing the initial image $\frac{20}{3}$ cm to the right of the first lens.

Given that the second lens is positioned 2 cm to the right, the light rays are intercepted at a distance of $\frac{20}{3} - 2 = \frac{14}{3}$ cm from the second lens, which now acts as the object distance $u_2$. Utilizing the lens formula for the second lens with a focal length of $-5$ cm yields $\frac{1}{v_2} - \frac{1}{\frac{14}{3}} = \frac{1}{-5}$. Solving this calculation results in $\frac{1}{v_2} = \frac{3}{14} - \frac{1}{5} = \frac{1}{70}$, meaning $v_2 = 70$ cm. Because the calculated image distance is positive, the resulting image is real and forms 70 cm to the right of the second lens.

Total internal reflection occurs at an interface when the angle of incidence exceeds the critical angle $C$, which is defined by the relationship $\sin C = \frac{\mu_1}{\mu_2}$. For a ray entering a cube at face AB, it refracts at an angle $r$, governed by Snell's Law as $\mu_1 \sin \theta = \mu_2 \sin r$. Because the faces of the cube are perpendicular to each other, the geometry of the system dictates that the angle of incidence at the second face, BC, is the complement of the refraction angle at the first face, given by $90^\circ - r$.

For total internal reflection to occur at face BC, this incidence angle must satisfy the inequality $90^\circ - r > C$. Rearranging this condition to $r < 90^\circ - C$ allows us to compare the sines of both sides, resulting in $\sin r < \sin(90^\circ - C)$, which simplifies to $\sin r < \cos C$. Since $\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \left(\frac{\mu_1}{\mu_2}\right)^2}$, we can express this limit as $\sin r < \frac{\sqrt{\mu_2^2 - \mu_1^2}}{\mu_2}$. Substituting the expression for $\sin r$ from Snell's Law into this inequality yields $\frac{\mu_1 \sin \theta}{\mu_2} < \frac{\sqrt{\mu_2^2 - \mu_1^2}}{\mu_2}$. Solving this for $\theta$ leads to $\sin \theta < \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$, which confirms that the angle of incidence must be less than $\sin^{-1} \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$.

The lens maker's formula, $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, allows us to relate focal length to the refractive index and radius of curvature of lens surfaces. For a plano-convex lens with index $\mu_1$, the focal length follows $\frac{1}{f_1} = \frac{\mu_1 - 1}{R}$, whereas for a plano-concave lens with index $\mu_2$, the magnitude of the focal length is $\frac{1}{f_2} = \frac{\mu_2 - 1}{R}$. Given the condition $f_1 = 2f_2$, we equate $\frac{\mu_1 - 1}{R} = \frac{1}{2} \left( \frac{\mu_2 - 1}{R} \right)$, which simplifies to $2(\mu_1 - 1) = \mu_2 - 1$. Solving this algebraic expression leads to $2\mu_1 - \mu_2 = 1$.

For an equilateral prism, the apex angle $A$ is $60^\circ$. During minimum deviation, light travels symmetrically through the prism, which forces the internal angle of refraction to be half the apex angle, resulting in $r = \frac{A}{2} = 30^\circ$. Applying Snell's law at the incident surface, $\sin i = \mu \sin r$, we substitute the refractive index $\mu = \sqrt{3}$ and the refraction angle $r = 30^\circ$ to obtain $\sin i = \sqrt{3} \cdot \sin 30^\circ = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$. This value confirms that the angle of incidence is $60^\circ$.

We first determine where the convex lens forms an image by using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. With an object placed 30 cm to the left of the 20 cm focal length lens, the light converges at a point 60 cm to the right of the lens. Because the mirror is positioned 80 cm from the lens, this convergence point lies 20 cm in front of the concave mirror. For the image to remain in the same position when the mirror is present, the mirror must reflect the incoming light rays back along their original path, which only occurs if the light strikes the mirror surface normally, directed toward its center of curvature. Consequently, the radius of curvature must be 20 cm, implying that the focal length of the mirror is 10 cm.

A concave mirror creates a virtual image only when an object is positioned between its pole and its focus. Given that the focal length of this mirror is 10 cm, any object placed at a distance less than this focal length will form a virtual image. Therefore, the maximum distance of the object for which the mirror produces a virtual image is 10 cm.

The angle of minimum deviation for a thin prism is determined by the relationship $D_m = (\mu - 1)A$, where $A$ represents the prism angle and $\mu$ is the refractive index. For crown glass, the refractive index decreases as the wavelength of incident light increases. Since the angle of minimum deviation $D_m$ depends directly on this refractive index, a rise in wavelength consequently forces a reduction in the value of $D_m$, resulting in a downward-sloping curve.