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    Geometrical Optics

    Flashcards for topic Geometrical Optics

    Intermediate105 cardsGeneral

    Preview Cards

    Card 1

    Front

    A light ray passes from medium 1 (μ₁) to medium 2 (μ₂) through a spherical interface with radius R. Derive the relationship between object distance u, image distance v, and radius R.

    Back

    The formula for refraction at a spherical interface is:

    μ1u+μ2v=μ2−μ1R\frac{\mu_1}{u} + \frac{\mu_2}{v} = \frac{\mu_2 - \mu_1}{R}uμ1​​+vμ2​​=Rμ2​−μ1​​

    Derivation steps:

    1. Consider a paraxial ray from object striking surface at angle i
    2. Apply Snell's law: μ₁sin(i) = μ₂sin(r)
    3. For small angles: sin(i) ≈ i and sin(r) ≈ r
    4. Use geometric relationships between angles and distances
    5. Combine and simplify equations

    Sign convention:

    • Distances measured along the direction of incident light are positive
    • Distances measured opposite to incident light are negative
    • R positive when center of curvature is in second medium
    • R negative when center of curvature is in first medium

    This equation works for all spherical interfaces when proper signs are used and is fundamental to lens design and optical systems analysis.

    Card 2

    Front

    When a point source S is placed in front of a plane mirror, how can you prove that all reflected rays, when extended backward, will meet at a single point I? What properties must this point I have?

    Back

    Proof that all reflected rays meet at a single point I:

    1. For any ray from source S striking the mirror at point P, the angle of incidence equals the angle of reflection
    2. The normal at P bisects the angle between the incident and reflected rays
    3. This creates congruent triangles on either side of the mirror
    4. Therefore, point I must be:
      • Located directly opposite to S
      • At the same perpendicular distance behind the mirror as S is in front
      • On a line perpendicular to the mirror passing through S

    The point I has these essential properties:

    • Forms a perpendicular with source S to the mirror
    • Distance from mirror = distance of S from mirror
    • Acts as the apparent origin of all reflected rays
    • Serves as a virtual image (rays don't actually pass through it)

    This demonstrates why a plane mirror forms a single, well-defined virtual image regardless of the observer's position.

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    Card 3

    Front

    When tracking a light ray through a prism, how do you determine the mathematical relationship between the angles of incidence, refraction, and the total deviation?

    Back

    The total deviation (δ) of a light ray passing through a prism relates to the various angles as follows:

    1. For any ray path through a prism:

      • δ = (i + i') - A Where:
      • i = angle of incidence at first surface
      • i' = angle of emergence at second surface
      • A = angle of the prism
      • δ = total deviation angle
    2. At the refracting surfaces:

      • The angles of refraction (r and r') inside the prism must satisfy: r + r' = A
      • This comes from the geometry of the prism
    3. Using Snell's Law at each surface:

      • sin(i)/sin(r) = μ (for first surface)
      • sin(i')/sin(r') = μ (for second surface)

    Note: The deviation is always minimized when the ray passes symmetrically through the prism (i = i' and r = r').

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    Card 4

    Front

    Derive the lens formula from first principles

    Back

    Starting with refraction at two spherical surfaces:

    1. For first surface: μ2v1−μ1u=μ2−μ1R1\frac{\mu_2}{v_1} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1}v1​μ2​​−uμ1​​=R1​μ2​−μ1​​
    2. For second surface: μ1v−μ2v1=μ1−μ2R2\frac{\mu_1}{v} - \frac{\mu_2}{v_1} = \frac{\mu_1 - \mu_2}{R_2}vμ1​​−v1​μ2​​=R2​μ1​−μ2​​
    3. Adding these equations: μ1v−μ1u=(μ2−μ1)(1R1−1R2)\frac{\mu_1}{v} - \frac{\mu_1}{u} = (\mu_2 - \mu_1)(\frac{1}{R_1} - \frac{1}{R_2})vμ1​​−uμ1​​=(μ2​−μ1​)(R1​1​−R2​1​)
    4. For a lens in air (μ1=1\mu_1 = 1μ1​=1, μ2=μ\mu_2 = \muμ2​=μ): 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}v1​−u1​=f1​
    5. Where 1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})f1​=(μ−1)(R1​1​−R2​1​)
    Card 5

    Front

    Compare and contrast the first and second focal points of a lens, and explain how focal length relates to lens power

    Back

    First Focal Point (F₁):

    • Point where object must be placed to produce parallel emergent rays
    • For convergent lens: Real point on incident side
    • For divergent lens: Virtual point on emergent side

    Second Focal Point (F₂):

    • Point where parallel incident rays converge after refraction
    • For convergent lens: Real point on emergent side
    • For divergent lens: Virtual point on incident side

    Relationship between focal points:

    • For thin lens in homogeneous medium: |PF₁| = |PF₂|
    • Usually just referred to as "the focal length f"

    Lens Power:

    • P = 1/f (measured in diopters, m⁻¹)
    • Positive power = converging lens
    • Negative power = diverging lens
    • For lenses in contact: P = P₁ + P₂
    Card 6

    Front

    How is lateral magnification calculated when light refracts through a spherical boundary between two media with refractive indices μ₁ and μ₂?

    Back

    The lateral magnification is given by:

    m=μ1vμ2um = \frac{\mu_1 v}{\mu_2 u}m=μ2​uμ1​v​

    Where:

    • m = lateral magnification
    • u = object distance (negative for real objects)
    • v = image distance (positive for real images)
    • μ₁ = refractive index of the medium containing the object
    • μ₂ = refractive index of the medium containing the image

    This formula accounts for the fact that magnification in refraction depends not only on the ratio of distances but also on the ratio of refractive indices, unlike in reflection where only distances matter.

    Example: If light travels from air (μ₁=1) into glass (μ₂=1.5) and an object at 20 cm from the boundary forms an image at 40 cm inside the glass, the magnification would be m = (1×40)/(1.5×-20) = -1.33.

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    Card 7

    Front

    How does a convex lens form an image of an extended object perpendicular to the principal axis?

    Back

    A convex lens forms an image of an extended object through:

    1. Ray tracing using two principal rays:

      • Ray through optical center (P) travels undeviated
      • Ray parallel to principal axis passes through the second focal point (F₂)
    2. The image location is determined by where these rays intersect

    3. For an object OQ perpendicular to the principal axis:

      • The image O'Q' will also be perpendicular to the principal axis
      • The height ratio of image to object (h₂/h₁) gives the lateral magnification (m)
      • m = h₂/h₁ = v/u (where v is image distance, u is object distance)

    The process follows lens equation: 1/v - 1/u = 1/f

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    Card 8

    Front

    What is the mathematical relationship between a lens's aperture size, focal length, and the severity of spherical aberration?

    Back

    The severity of spherical aberration relates to lens parameters by:

    • Longitudinal spherical aberration (LSA) ≈ k × (D/f)² × f Where:
      • k is a shape factor depending on lens design
      • D is the lens diameter (aperture)
      • f is the focal length

    Key relationships:

    • Aberration increases with the square of the aperture diameter
    • Aberration is proportional to focal length for a given f-number (f/D ratio)
    • The f-number (f/D) is inversely related to aberration severity
    • For a given lens shape, doubling the aperture increases aberration by factor of 4

    Practical impact:

    • Fast lenses (small f-number) exhibit more spherical aberration
    • Circle of least confusion size determines practical resolution limit
    • Stopping down (decreasing aperture) rapidly reduces aberration
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    Card 9

    Front

    What optical principle explains why a planoconvex lens performs better when its curved surface faces incident parallel light, and how might this principle be applied to other optical designs?

    Back

    • Principle: Distributing optical power across multiple surfaces minimizes aberrations

    • Physical explanation:

      • When refraction is divided between surfaces, ray path angles are smaller at each interface
      • Smaller angles produce less deviation from paraxial approximation
      • Total aberration is roughly proportional to cube of refraction angle
    • Applications to other designs:

      • Achromatic doublets distribute power between elements
      • Camera lenses use multiple elements rather than single powerful lens
      • Microscope objectives use multiple elements with distributed power
      • Telescope designs benefit from compound lens arrangements
    • This principle is fundamental to designing high-performance optical systems with minimal aberrations

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    Card 10

    Front

    Explain the minimum deviation through a prism: What conditions create it, how is it related to the refractive index (μ), and how would you derive the formula and apply it experimentally?

    Back

    Minimum Deviation in a Prism

    Conditions for Minimum Deviation:

    • Ray passes symmetrically through the prism
    • Angle of incidence equals angle of emergence (i = i′)
    • Angle of refraction at both surfaces is equal (r = r′)

    Key Formula:

    μ=sin⁡(A+δm2)sin⁡(A2)\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}μ=sin(2A​)sin(2A+δm​​)​

    Where:

    • μ = refractive index of prism material
    • A = angle of the prism
    • δₘ = angle of minimum deviation

    Derivation Steps:

    1. For symmetric passage: i = i′ and r = r′
    2. From geometry: A = r + r′ = 2r (due to symmetry)
      • Therefore: r = A/2
    3. From exterior angle: δₘ = i + i′ - A = 2i - A (due to symmetry)
      • Therefore: i = (A + δₘ)/2
    4. Apply Snell's law: μ = sin(i)/sin(r)
    5. Substitute: μ = sin((A + δₘ)/2)/sin(A/2)

    Small Angle Approximation:

    For small angles, the formula simplifies to: δm=(μ−1)A\delta_m = (μ-1)Aδm​=(μ−1)A

    Experimental Application:

    • Measure the angle of minimum deviation (δₘ) for a prism of known angle A
    • Use the formula to calculate the refractive index μ of the prism material
    • Example: For a prism with A = 60° and measured δₘ = 30°, μ = 1.5

    Note: Minimum deviation occurs only at one specific angle of incidence, making it a precise method for determining refractive indices.

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