A hemispherical vessel is completely filled with a liquid of refractive index μ$\mu$ . A small coin is kept at the lowest point (O) of the vessel as shown in figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point E (at the level of the vessel) is

Related Formula

Magnification relation for spherical mirrors:

Mirror equation:

Core Logic

Given m = -(1)/(3)$m = -\frac{1}{3}$:

Since magnification is negative, a real inverted image is formed on the same side as the object in a concave mirror layout configuration.

Step 1: Formulate Position Distances

The distance between the object and image is given as 30~cm$30\mathrm{~cm}$:

This gives u = 3(15) = 45~cm$u = 3(15) = 45\mathrm{~cm}$.

Step 2: Solve for Focal Length and x

Apply standard sign conventions (u = -45~cm$u = -45\mathrm{~cm}$, v = -15~cm$v = -15\mathrm{~cm}$):

Matching this with the prompt pattern form (x)/(4)$\frac{x}{4}$ yields: x = 45$x = 45$

Pattern Recognition

Real inverted diminished images (|m| < 1$|m| < 1$) mean the image forms closer to the mirror surface than the object, situated between the focal center F$F$ and center of curvature C$C$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

m = -(v)/(u)$$m = -\frac{v}{u}$$ [cite: 700]

(1)/(v) + (1)/(u) = (1)/(f)$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ [cite: 707]

Core Logic

Given magnification magnitude |m| = (1)/(4)$|m| = \frac{1}{4}$[cite: 113]. Assuming a real image formed by a concave mirror: [cite: 701]

(v)/(u) = (1)/(4) u = 4v$$\frac{v}{u} = \frac{1}{4} \implies u = 4v$$ [cite: 701]

The distance between the object and the image is given as 40 cm$40\ \text{cm}$ [cite: 114, 116]:

Applying mirror sign conventions (u = -(160)/(3)$u = -\frac{160}{3}$, v = -(40)/(3)$v = -\frac{40}{3}$): [cite: 701, 704]

f = -(160)/(15) ≈ -10.67 cm$$f = -\frac{160}{15} \approx -10.67\ \text{cm}$$ [cite: 712]

Rounding to the matching options choice gives 10.7 cm$10.7\ \text{cm}$[cite: 118, 712].

Pattern Recognition

Pay attention to sign conventions in mirror systems. A smaller real image formed by a concave mirror sits between the focus and center of curvature, resulting in u > v$u > v$ configurations.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Core Logic

Refractive index represents the ratio of the speed of light in vacuum to its speed in a given medium[cite: 719]. Glass slows light down more than air does, hence μglass ≈ 1.5 > μₐᵢᵣ ≈ 1.0$\mu_{\text{glass}} \approx 1.5 > \mu_{\text{air}} \approx 1.0$, which makes Assertion (A) correct[cite: 130]. However, optical density is defined by a medium's capacity to refract light and is completely conceptually distinct from inertial mass density (mass per unit volume)[cite: 719]. For example, turpentine has a lower mass density than water but possesses a higher optical density and refractive index. Therefore, Reason (R) is fundamentally incorrect[cite: 722].

Pattern Recognition

Optical density vs mass density is a signature conceptual trick in refraction theory[cite: 719]. They share the word 'density' but have entirely different physical meanings and no fixed mathematical proportionality[cite: 719, 722].

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

θc = (μ₁)/(μ₂)$$\sin\theta_c = \frac{\mu_1}{\mu_2}$$ [cite: 810]

μ₁ θ = μ₂ r$$\mu_1 \sin\theta = \mu_2 \sin r$$ [cite: 807]

Core Logic

From the boundary geometry at the top interface, the angle of refraction r$r$ at the first surface satisfies: [cite: 806]

r + θc = 90° r = 90° - θc$$r + \theta_c = 90^{\circ} \implies r = 90^{\circ} - \theta_c$$ [cite: 806]

Applying Snell's law at the first entry interface: [cite: 170, 807]

μ₁ θ = μ₂ r = μ₂ (90° - θc) = μ₂ θc$$\mu_1 \sin\theta = \mu_2 \sin r = \mu_2 \sin(90^{\circ} - \theta_c) = \mu_2 \cos\theta_c$$ [cite: 172, 173, 807, 809]

Since θc = (μ₁)/(μ₂) = (1.0)/(1.25) = (4)/(5)$\sin\theta_c = \frac{\mu_1}{\mu_2} = \frac{1.0}{1.25} = \frac{4}{5}$, we have θc = √(1 - ((4)/(5))²) = (3)/(5)$\cos\theta_c = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$[cite: 169, 810, 811]. Substituting back into the expression: [cite: 811]

1.0 · θ = 1.25 × (3)/(5) = (5)/(4) × (3)/(5) = (3)/(4)$$1.0 \cdot \sin\theta = 1.25 \times \frac{3}{5} = \frac{5}{4} \times \frac{3}{5} = \frac{3}{4}$$ [cite: 169, 811]

θ = ⁻¹((3)/(4))$$\theta = \sin^{-1}\left(\frac{3}{4}\right)$$ [cite: 811]

Pattern Recognition

Maximum angle at the entry face ensures minimum angle of incidence at the subsequent wall[cite: 806, 807]. Setting that exact internal angle equal to the critical threshold condition values solves for the operational scanning range edge directly[cite: 808].

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

Areal Magnification:

since object distance u = -18 km$u = -18\ \mathrm{km}$ is vastly larger than f$f$.

Core Logic

Given parameters:

• Object height distance, H = 18 km = 18 × 10³ m$H = 18\ \mathrm{km} = 18 \times 10^3\ \mathrm{m}$
• Film size area, Aimage = 2 cm × 2 cm = 4 cm² = 4 × 10⁻⁴ m²$A_{\text{image}} = 2\ \mathrm{cm} \times 2\ \mathrm{cm} = 4\ \mathrm{cm^2} = 4 \times 10^{-4}\ \mathrm{m^2}$
• Landscape area, Aobject = 400 km² = 400 × 10⁶ m²$A_{\text{object}} = 400\ \mathrm{km^2} = 400 \times 10^6\ \mathrm{m^2}$

Linear magnification factor:

Using the simple pinhole/thin lens perspective ratio:

Pattern Recognition

For aerial satellite imaging contexts where u gg f$u \gg f$, linear sizing scales directly as film sideground side = (f)/(H)$\frac{\text{film side}}{\text{ground side}} = \frac{f}{H}$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments