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#optics

63 public questions tagged with this topic.

PHYSICS

A compound microscope has an objective of focal length 2 cm and eyepiece of focal length 5 cm with a tube length of 18 c

Given: A compound microscope has an objective of focal length 2 cm and eyepiece of focal length 5 cm with a tube length of 18 cm . What is the magnification at infinity? These values define the system as per NCERT data. Formula: Objective magnification: m_o = L/f_o = 18/2 = 9. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Eyepiece magnification: m_e = D/f_e = 25/5 = 5 . Total magnification: m = m_o × m_e = 9 × 5 = 45 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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CHEMISTRY

Calculate the wavenumber of light with wavelength 5800 Å . (1 Å = 10 ⁻¹⁰ m)

Given: Calculate the wavenumber of light with wavelength 5800 Å . (1 Å = 10 ⁻¹⁰ m) These values define the system as per NCERT data. Formula: Wavenumber barnu = 1/lambda. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 5800 Å = 5800 × 10⁻¹⁰ m = 5.8 × 10⁻⁷ m . barnu = frac15.8 × 10⁻⁷= 1.724 × __10POW₆__m^{-1 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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PHYSICS

What is the angular position of the first minimum in a single-slit diffraction pattern if the slit width is 5.0 μm and

Given: What is the angular position of the first minimum in a single-slit diffraction pattern if the slit width is 5.0 μm and the wavelength is 500 nm ? These values define the system as per NCERT data. Formula: First minimum occurs at sin θ = lambda/a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 5.0 × 10⁻⁷ m, a = 5.0 × 10⁻⁶ m . sin θ = frac5.0 × 10⁻⁷⁵.0 × 10⁻⁶= 0.1, θ = sin^{-1(0.1) approx 5.7° . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Ray Optics and Optical Instruments and Wave Optics, Topic: Refraction, lenses and interference/diffraction.

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PHYSICS

In a single-slit diffraction experiment, if the slit width is 2.5 μm and the wavelength is 500 nm, what is the angle of

Given: In a single-slit diffraction experiment, if the slit width is 2.5 μm and the wavelength is 500 nm, what is the angle of the first minimum? These values define the system as per NCERT data. Formula: First minimum occurs at sin θ = lambda/a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 500 nm = 5.0 × 10⁻⁷ m, a = 2.5 μm = 2.5 × 10⁻⁶ m . sin θ = frac5.0 × 10⁻⁷².5 × 10⁻⁶= 0.2, so θ = sin^{-1(0.2) approx 11.5° . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Ray Optics and Optical Instruments and Wave Optics, Topic: Refraction, lenses and interference/diffraction.

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PHYSICS

A convex mirror has a radius of curvature of 50 cm . An object is placed 25 cm from it. What is the image distance?

Given: A convex mirror has a radius of curvature of 50 cm . An object is placed 25 cm from it. What is the image distance? Formula: Focal length: f = R/2 = 50/2 = 25 cm (positive for convex). Substitution & Calculation: Object distance: u = -25 cm . Mirror equation: 1/v + 1/u = 1/f . 1/v + 1/-25 = 1/25 Rightarrow 1/v = 1/25 + 1/25 = 2/25 . v = 25/2 = 12.5 cm (virtual image). Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A prism of refracting angle 30° and refractive index 1.6 produces what minimum deviation?

Given: A prism of refracting angle 30° and refractive index 1.6 produces what minimum deviation? Formula: For a thin prism: D_m = (n - 1) A. Substitution & Calculation: n = 1.6, A = 30° . D_m = (1.6 - 1) × 30 = 0.6 × 30 = 18° . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

An object is at a depth of 19.95 cm in a medium with refractive index 1.5 . What is the apparent depth?

Given: An object is at a depth of 19.95 cm in a medium with refractive index 1.5 . What is the apparent depth? Formula: Apparent depth = fracreal depthn. Substitution & Calculation: Real depth = 19.95 cm, n = 1.5 . Apparent depth = 19.95/1.5 = 13.3 cm . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

In a single-slit diffraction pattern, what is the angular width of the ntral maximum if the slit width is 2.0 μm and th

Given: In a single-slit diffraction pattern, what is the angular width of the ntral maximum if the slit width is 2.0 μm and the wavelength is 400 nm ? These values define the system as per NCERT data. Formula: Angular width of the ntral maximum 2θ = 2lambda/a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 4.0 × 10⁻⁷ m, a = 2.0 × 10⁻⁶ m . sin θ = lambda/a = frac4.0 × 10⁻⁷².0 × 10⁻⁶= 0.2, θ = sin^{-1(0.2) approx 11.5°, 2θ approx 23° . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Ray Optics and Optical Instruments and Wave Optics, Topic: Refraction, lenses and interference/diffraction.

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PHYSICS

A prism of angle 45° has a refractive index of 1.5 . What is the angle of minimum deviation?

Given: A prism of angle 45° has a refractive index of 1.5 . What is the angle of minimum deviation? Formula: For a thin prism: D_m = (n - 1) A. Substitution & Calculation: n = 1.5, A = 45° . D_m = (1.5 - 1) × 45 = 0.5 × 45 = 22.5° . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A double convex lens of refractive index 1.55 has radii of curvature 30 cm and -30 cm . What is its focal length?

Given: A double convex lens of refractive index 1.55 has radii of curvature 30 cm and -30 cm . What is its focal length? These values define the system as per NCERT data. Formula: Lens maker’s formula: 1/f = (n - 1) ( 1/R_1 - 1/R_2 ). This is standard NCERT relation. Substitution & Calculation: n = 1.55, R_1 = 30 cm, R_2 = -30 cm . 1/f = (1.55 - 1) ( 1/30 - 1/-30 ) = 0.55 ( 1/30 + 1/30 ) = 0.55 × 2/30 = 1.1/30 . f = 30/1.1 approx 27.27 cm . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

What is the critical angle for a diamond ( n = 2.42 ) to water ( n = 1.33 ) interface?

Given: What is the critical angle for a diamond ( n = 2.42 ) to water ( n = 1.33 ) interface? Formula: Critical angle: sin i_c = n_2/n_1. Substitution & Calculation: Diamond ( n_1 = 2.42 ), water ( n_2 = 1.33 ). sin i_c = 1.33/2.42 approx 0.55 . i_c = sin^{-1(0.55) approx 33.4° . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

What is the critical angle for a dense flint glass ( n = 1.62 ) to air interface?

Given: What is the critical angle for a dense flint glass ( n = 1.62 ) to air interface? These values define the system as per NCERT data. Formula: Critical angle: sin i_c = n_2/n_1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Glass ( n_1 = 1.62 ), air ( n_2 = 1 ). sin i_c = 1/1.62 approx 0.617 . i_c = sin^{-1(0.617) approx 38.1° . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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