Question
- Spherical aberration in a thin lens is minimised if rays are bent equally by the two surfaces. If a planoconvex lens is used to form a real image of an object at infinity, which surface should face the object?
Answer
The curved surface should face the object (incoming parallel rays). This distributes the refraction more evenly between the two surfaces. If the flat side faces the object, all bending occurs at the curved surface on exit, maximising spherical aberration. With the curved side first, rays are partially refracted on entry and further refracted on exit, sharing the bending.
Question
- In attempting to discern distant details, people will sometimes squint. Why does this help?
Answer
Squinting reduces the effective aperture of the eye, blocking marginal rays that are most affected by aberrations and defocus. This increases the depth of field and reduces the blur circle on the retina, producing a sharper (though dimmer) image. It also reduces spherical aberration by using only the paraxial region of the lens.
Question
- For both converging and diverging lenses, discuss how the focal length for red light differs from that for violet light.
Answer
Due to dispersion, n is higher for violet (shorter λ) than red. From the lensmaker’s equation 1/f = (n-1)(1/R₁ ± 1/R₂), higher n → shorter f. So for both converging and diverging lenses, |f_violet| < |f_red|. Converging lens: violet focuses closer. Diverging lens: violet has a shorter (more negative) focal length, diverging more strongly. This is chromatic aberration.
Question
- Does the focal length of a lens depend on the fluid in which it is immersed? What about the focal length of a spherical mirror?
Answer
Lens: yes. The lensmaker’s equation depends on the relative refractive index (n_lens/n_medium - 1). Immersing in a fluid with higher n reduces the difference, increasing the focal length. If n_fluid = n_lens, the lens has no optical power. Mirror: no. Reflection follows the law of reflection regardless of the medium. f = R/2 depends only on geometry, not refractive index.
Question
- An underwater lens consists of a carefully shaped thin-walled plastic container filled with air. What shape should it have to be (a) converging (b) diverging?
Answer
The air pocket has lower n than surrounding water, so the lens behaviour is inverted compared to glass in air. (a) Converging: biconcave shape. The concave air-filled lens in water acts as a converging lens because the lower-n medium is shaped to converge rays (opposite of glass in air). (b) Diverging: biconvex shape. A convex air pocket in water diverges light.
Question
- A sharp image is located 88.0 mm behind a 75.0 mm focal-length converging lens. Find the object distance.
Answer
Thin lens equation: 1/do = 1/f - 1/di = 1/75.0 - 1/88.0 1/do = (88.0 - 75.0)/(75.0 × 88.0) = 13.0/6600 = 1/507.7 do ≈ 508 mm (50.8 cm) The object is far from the lens, consistent with the image being close to (but beyond) the focal point.
Question
- A diverging lens with f = -33.5 cm is placed 14.0 cm behind a converging lens with f = 20.0 cm. Where will an object at infinity be focused?
Answer
Object at infinity → converging lens forms image at its focal point: di₁ = 20.0 cm (behind lens 1). This image is 20.0 - 14.0 = 6.0 cm behind lens 2, so it acts as a virtual object for lens 2: do₂ = -6.0 cm (negative because it’s behind lens 2). 1/di₂ = 1/f₂ - 1/do₂ = 1/(-33.5) - 1/(-6.0) = -0.02985 + 0.1667 = 0.1368 di₂ ≈ 7.3 cm behind lens 2. The final image is real, ~7.3 cm behind the diverging lens.
Question
- Two 25.0 cm focal-length converging lenses are placed 16.5 cm apart. An object is placed 35.0 cm in front of one lens. Where will the final image be located? What is the total magnification?
Answer
Lens 1: 1/di₁ = 1/25.0 - 1/35.0 = (35-25)/(25×35) = 10/875 di₁ = 87.5 cm. This is beyond lens 2, so it’s a virtual object for lens 2. do₂ = -(87.5 - 16.5) = -71.0 cm 1/di₂ = 1/25.0 - 1/(-71.0) = 0.04 + 0.01408 = 0.05408 di₂ ≈ 18.5 cm behind lens 2. m₁ = -di₁/do₁ = -87.5/35.0 = -2.50 m₂ = -di₂/do₂ = -18.5/(-71.0) = -0.260 m_total = m₁ × m₂ = (-2.50)(-0.260) = 0.651 Final image is real, 18.5 cm behind lens 2, upright relative to the object, magnification ≈ 0.65.
Question
- A microscope uses an eyepiece with fe = 1.50 cm. Barrel length is 17.5 cm and fo = 0.65 cm. Final image at infinity. What is the magnification?
Answer
M = -(L/fo)(N/fe) where L = barrel length - fe - fo ≈ 17.5 - 1.50 - 0.65 = 15.35 cm, N = 25 cm (near point). M = -(15.35/0.65)(25/1.50) = -23.6 × 16.7 ≈ -394 Magnification ≈ 390×. Negative sign indicates inverted image.
Question
- Planoconvex lens: R = 15.3 cm (one flat surface). Object at 66.0 cm. n_red = 1.5106, n_yellow = 1.5226. Find image locations for red and yellow light.
Answer
Flat surface: R₁ = ∞. Curved surface: R₂ = -15.3 cm (using convention where centre of curvature is on the incoming side). 1/f = (n-1)(1/R₁ - 1/R₂) = (n-1)(0 - 1/(-15.3)) = (n-1)/15.3 Red: 1/f_red = 0.5106/15.3 = 0.03337 → f_red = 29.97 cm Yellow: 1/f_yellow = 0.5226/15.3 = 0.03416 → f_yellow = 29.27 cm Red image: 1/di = 1/29.97 - 1/66.0 = 0.03337 - 0.01515 = 0.01822 → di_red ≈ 54.9 cm Yellow image: 1/di = 1/29.27 - 1/66.0 = 0.03416 - 0.01515 = 0.01901 → di_yellow ≈ 52.6 cm The yellow image forms ~2.3 cm closer — this separation is chromatic aberration.
Question
- Achromatic lens: f₁ = -28 cm, f₂ = +25 cm in contact. (a) Converging or diverging? (b) Net focal length?
Answer
1/f = 1/f₁ + 1/f₂ = 1/(-28) + 1/25 = -0.03571 + 0.04000 = 0.00429 f ≈ 233 cm (a) Converging (positive f). (b) f ≈ 233 cm. The combination is weakly converging. An achromatic doublet uses a diverging element of different glass (different dispersion) to cancel chromatic aberration while maintaining net positive power.
Question
- Which colour of visible light would give the best resolution in a microscope?
Answer
Violet/blue. Resolution (minimum resolvable distance) = 0.61λ/NA. Shorter wavelength → smaller resolvable features. Violet light (~400nm) gives the best resolution of visible colours.
Question
- Can visible light be used to “see” an atom (~10⁻¹⁰ m) with an optical microscope?
Answer
No. Visible light wavelengths are 400-700nm (4-7 × 10⁻⁷ m), roughly 1000× larger than an atom. The diffraction limit prevents resolving features much smaller than the wavelength. You need electron microscopy or scanning probe techniques (STM/AFM) for atomic resolution.
Question
- For diffraction by a single slit, what is the effect of increasing (a) the slit width, (b) the wavelength?
Answer
First minimum at sinθ = λ/a. (a) Increasing slit width a: diffraction pattern narrows (central maximum becomes narrower). In the limit a → ∞, no diffraction. (b) Increasing wavelength λ: diffraction pattern widens (central maximum becomes broader). More spreading for longer wavelengths.
Question
- Describe the single-slit diffraction pattern when white light falls on a slit of width (a) 50 nm, (b) 50,000 nm.
Answer
(a) 50 nm: slit is much smaller than visible wavelengths (400-700nm). sinθ = λ/a > 1 for all colours, so there are no minima — light diffracts broadly in all directions. Essentially uniform illumination across the screen. (b) 50,000 nm (50 µm): slit is ~100× larger than wavelengths. Very narrow central maximum with many closely spaced fringes. Each colour diffracts by a slightly different amount, producing coloured fringes at the edges of the central white peak.
Question
- Explain why diffraction patterns are more difficult to observe with an extended light source than a point source. Compare monochromatic to white light.
Answer
Extended source: each point on the source produces its own diffraction pattern, shifted slightly. These overlap incoherently, washing out the fringe pattern. A point source gives a single sharp pattern. Monochromatic: sharp, well-defined minima and maxima at fixed positions. White light: each wavelength has minima at different angles, so only the central maximum is truly bright and white. Higher-order fringes blur into coloured bands and eventually overlap, reducing contrast.
Question
- What happens to the diffraction pattern of a single slit if the apparatus is immersed in (a) water, (b) vacuum?
Answer
(a) Water (n ≈ 1.33): the effective wavelength decreases (λ_medium = λ₀/n). The diffraction pattern narrows — central maximum becomes smaller, improving resolution. (b) Vacuum: wavelength is marginally longer than in air (n_air ≈ 1.0003), so the pattern is essentially unchanged — very slightly wider.
Question
- Does diffraction limit the resolution of images formed by (a) spherical mirrors, (b) plane mirrors?
Answer
(a) Spherical mirrors: yes. They have a finite aperture, so diffraction limits resolution just as with lenses. The Rayleigh criterion applies: θ_min = 1.22λ/D. (b) Plane mirrors: no (in the conventional sense). Plane mirrors don’t form images by converging light through an aperture — they simply reflect. The image quality is limited by the mirror’s flatness and the imaging system viewing the reflection, not by diffraction at the mirror itself.
Question
- Do diffraction effects occur for virtual as well as real images?
Answer
Yes. Diffraction occurs whenever light passes through any aperture (lens, mirror, slit). Since virtual images are formed by the same optical elements that impose diffraction, the resolution of virtual images is equally limited. The light still passes through a finite aperture regardless of where the image appears.
Question
- Violet light (λ = 415 nm) creates a central diffraction peak 8.20 cm wide on a screen 2.85 m away. How wide is the slit?
Answer
Half-width of central max = 4.10 cm = 0.0410 m. First minimum: sinθ = λ/a, and tanθ ≈ sinθ = y/L = 0.0410/2.85 = 0.01439 a = λ/sinθ = 415×10⁻⁹/0.01439 = 2.88×10⁻⁵ m ≈ 28.8 µm
Question
- Give reasons for the use of reflecting mirrors instead of lenses in photolithography.
Answer
Mirrors have no chromatic aberration (no refraction, so no wavelength-dependent focusing). This is critical for EUV lithography (13.5nm) where no practical transmissive materials exist. Mirrors also avoid absorption losses in lens materials at short wavelengths, can be designed with fewer aberrations for wide fields, and don’t suffer from radiation damage that degrades lens materials over time.
Question
- What primarily limits the resolution of optical lithography? A) The speed of the stepper motor B) The quality of the semiconductor wafer C) The cleanliness of the manufacturing environment D) The wavelength of the light source used E) The temperature of the processing environment
Answer
D Resolution is fundamentally diffraction-limited: R = k₁λ/NA. Shorter wavelength → smaller features. This is why lithography has progressed from g-line (436nm) → i-line (365nm) → DUV (248nm, 193nm) → EUV (13.5nm).
Question
- What advanced technique is used in optical lithography to improve resolution beyond the diffraction limit of light? A) Electron beam lithography B) X-ray lithography C) Chemical vapour deposition D) Immersion lithography E) Multilayer photomasks
Answer
D Immersion lithography places a high-n fluid (typically water, n=1.44) between the final lens element and the wafer. This increases the effective NA (NA = n·sinθ), reducing the minimum resolvable feature size. 193nm immersion achieves NA > 1.0, enabling sub-40nm features.