As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.

If refractive index of the material of prism is √(2)$\sqrt{2}$, the angle θ$\theta$ of prism 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