Lenses and diffraction

A projector lens forms a picture on a screen, while a magnifying glass shows enlarged letters where a screen would not help. Narrow slits then send light in several directions. Let us find out which questions rays can answer and which need the phase of a wave.

Catch the light from a lens on a sheet of paper. If rays from one point actually converge, its real image appears on the paper. You can also see a point through a magnifying glass even when the outgoing rays diverge. Only their backward extensions then intersect, forming a virtual image: a screen placed at the apparent image position will not show it. The model distinguishes the actual light path from a geometrical extension, not a genuine visual experience from a false one.

To keep the construction simple, we neglect the lens thickness compared with the object and image distances. This is the thin-lens model. A beam parallel to the axis converges at the principal focus of a converging lens. A diverging lens spreads it as if it came from a virtual focus. Focal length is a property of the lens in a given surrounding medium. The image of a nearby object need not be at the focus.

Write F for focal length, d for object distance and f for image distance, all in metres. For a real object, d>0. A converging lens has F>0 and a diverging lens F<0. A real image has f>0, a virtual image f<0. These signs encode geometry under the convention we have chosen. Negative f does not mean a negative physical size, so we will use an absolute value when calculating image size.

For a thin lens and rays close to the axis, geometry gives 1/F=1/d+1/f. For a very distant object, 1/d is nearly zero and f≈F, as expected for a parallel beam. At d=F, 1/f=0: the outgoing rays are parallel and no image forms at a finite distance. Move a real object inside the focus of a converging lens and 1/f becomes negative, giving a virtual image. Insert all distances with their signs and in the same unit.

Check the answer using two rays from the top of the object. A ray through the optical centre of a thin converging lens is almost undeviated; one parallel to the axis passes through the far focus after the lens. The intersection of the rays or their extensions gives the image position and orientation. Imagine moving the object towards F: the real image moves farther away. Once the object passes inside the focus, the image is virtual and on the object side. The diagram helps check the sign of the answer.

A short focal length means stronger bending of a parallel beam, so optical power is defined as D=1/F. With F in metres, D is in dioptres: 1\,\text{дптр}=1\,\text{м}^{-1}. Halving |F| doubles |D| without changing the sign: positive for a converging lens and negative for a diverging lens. If the focus is very far away, the lens has a weak effect and D approaches zero. Convert centimetres to metres first.

The ray through the centre forms similar triangles on the object and image sides. This gives the size magnification \Gamma=H/h=|f/d|. H and h are positive sizes in the same units, and Γ has no units. For Γ>1 the image is larger; for Γ<1 it is smaller. In this geometry, doubling |f| at unchanged d doubles the size. A positive Γ does not mean an upright image: inversion is determined by the ray arrangement, not by the absolute value of the ratio.

Near narrow slits, a ray diagram is no longer enough: diffraction spreads light in different directions. A grating has many equally illuminated slits whose waves overlap. Where their phases align, the fields reinforce one another and bright principal maxima appear. Light does not choose a single slit and then turn by a special rule. The final pattern comes from the combined wave contributions of the slits.

For a grating, d now means the spacing between neighbouring slits in metres, not object distance. At normal incidence, waves travelling in a direction φ from neighbouring slits cover different distances: the difference is d sin φ. Measure φ from the normal to the grating. Along the normal the path difference is zero. Geometry connects a change of direction to a phase delay, which we already know how to calculate from wavelength.

Principal maxima occur when the path difference is a whole number of wavelengths: d\sin\varphi=k\lambda. Here k=0, ±1, ±2… is the dimensionless order and λ is in metres. For k=0 the angle is zero for every wavelength. Opposite signs of k give symmetric directions. The formula assumes monochromatic light, normal incidence and an ideal periodic grating. For mixed light, apply it to each component separately.

How many directions are possible? Since |sin φ|≤1, we require |kλ|≤d. A result of sin φ=1,3 means that this order does not exist; it is not an invitation to alter the number. At the same nonzero order, increasing λ increases the magnitude of the angle, while decreasing the spacing d spreads the maxima farther apart. This is why a grating separates colours. The possible orders are physically limited even though k in the formula looks like any integer.

An object is 0.30 m in front of a converging lens with F=0,10 м. Find the image distance and size magnification.

For a real object, d = 0,30 м is positive; for a converging lens, F = 0,10 м is positive. The object is beyond 2F = 0,20 м, so we expect a smaller real image.

1/f = 1/F − 1/d = 1/0,10 − 1/0,30 = 10 − 3,33… = 6,66… м⁻¹. Therefore f = 0,15 м.

Γ = |f/d| = 0,15/0,30 = 0,5: the linear size is halved.

Positive f and a position between F and 2F agree with the intersection of actual rays. The ray diagram also shows inversion, which a positive Γ alone cannot tell us.

f = 0,15 м, Γ = 0,5. The image is real, inverted and half the object size. Place the screen 15 cm beyond the lens.

Distance signs describe image geometry. Check them using actual rays and their extensions.

Optical power describes a lens through reciprocal focal length. Size magnification does not encode image inversion.

Principal grating maxima require a whole-wavelength path difference and an allowed angle: |kλ|≤d.

Next we will explore how measurements of time and length depend on the reference frame.