brodidit for students
Home / Notes / Class 11 / Physics / Optical instruments
Optical instruments: Class 11 Physics
Refraction and Snell's law, total internal reflection and optical fibres, lens formula and power, the human eye and its defects, the simple and compound microscope, and the astronomical telescope.
Key points
Refraction is the bending of light when it passes from one transparent medium to another. The refractive index of a medium is n = c/v, the ratio of the speed of light in vacuum to its speed in the medium. Snell's law says that n₁ sin θ₁ = n₂ sin θ₂, where the angles are measured from the normal. Light bends towards the normal when it enters a denser medium (larger n) and away from the normal when it enters a rarer medium. For water n is about 1.33, and for glass it is about 1.5.
Total internal reflection happens when light travels from a denser to a rarer medium and the angle of incidence is more than the critical angle C. At the critical angle the refracted ray grazes the surface (angle of refraction 90°), so sin C = 1/n for light going from the medium into air. Two conditions are needed: light must go from a denser to a rarer medium, and the angle of incidence must exceed C. For glass of n = 1.5, C is about 42°.
Applications of total internal reflection:
- Optical fibres: light is trapped in a thin glass core by repeated total internal reflections, and travels long distances with little loss. They are used in telephone and internet links and in the endoscope in medicine.
- Totally reflecting prisms in binoculars and periscopes, which turn light by 90° or 180°.
- The sparkle of a diamond, whose high refractive index (about 2.4) gives it a small critical angle.
- Mirages, caused by the total internal reflection of light in layers of hot air near the ground.
A lens is a piece of transparent material with curved surfaces. A convex (converging) lens is thicker in the middle, and a concave (diverging) lens is thinner in the middle. The principal focus F is the point where rays parallel to the axis meet (or appear to come from), and the focal length f is its distance from the lens. The lens formula is 1/f = 1/p + 1/q, where p is the object distance and q is the image distance (real is positive, in this convention). The magnification is m = −q/p = image height / object height.
The power of a lens is P = 1/f, with f in metres. Its unit is the dioptre (D). A convex lens has a positive power and a concave lens a negative power. A lens of focal length 50 cm has a power of 2 D. Lenses in contact add their powers: P = P₁ + P₂.
The human eye forms a real, inverted image on the retina using the cornea and the lens. The ciliary muscles change the focal length of the lens so that objects at different distances are in focus. This is called accommodation. The near point of a normal eye is about 25 cm and the far point is at infinity. The iris controls the size of the pupil, and so the amount of light entering the eye.
Defects of vision and their correction:
- Short sight (myopia): distant objects are blurred because the image forms in front of the retina. It is corrected with a concave lens.
- Long sight (hypermetropia): near objects are blurred because the image would form behind the retina. It is corrected with a convex lens.
- Astigmatism: the cornea is not equally curved in all directions. It is corrected with a cylindrical lens.
- Presbyopia: with age the lens loses its ability to accommodate, and bifocal lenses help.
The angular magnification (magnifying power) M of an optical instrument is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye, when it is placed at the near point (d = 25 cm) with the unaided eye. A simple microscope is a single convex lens of short focal length used as a magnifying glass, with the object placed inside the focal length. When the image is at the near point, M = 1 + d/f. When the image is at infinity (the relaxed eye), M = d/f.
A compound microscope has two convex lenses: an objective of very short focal length and an eyepiece of a longer focal length. The object is placed just beyond the focal point of the objective, which forms a real, magnified image. This image lies within the focal length of the eyepiece, which acts as a magnifier and gives a final virtual, inverted and highly magnified image. The total magnification is the product of the two: M = M_o × M_e. With the final image at the near point, M_e = 1 + d/f_e. It is used to look at cells and small organisms.
An astronomical telescope views distant objects such as stars and planets. Its objective has a large aperture and a long focal length f_o, and the eyepiece has a short focal length f_e. In normal adjustment the final image is at infinity, the lenses are separated by f_o + f_e, and the angular magnification is M = f_o/f_e. The final image is virtual and inverted. A large aperture collects more light, and the objective forms a real image at its focus. Large astronomical telescopes use curved mirrors instead of lenses (reflecting telescopes), because large mirrors are easier to make and support, and they do not show colour fringes. A terrestrial telescope adds an extra lens to give an upright image.
Study tip: State the sign convention you use for the lens formula at the start of the numerical. Learn the ray diagrams of the microscope and the telescope with labels for the objective, eyepiece, focal points and the final image. For defect questions, always mention where the image falls relative to the retina.
Definitions and flashcards
Say the meaning first, then open the card to check.
Refractive index
The ratio of the speed of light in vacuum to that in the medium.
Critical angle
The angle of incidence in the denser medium for which the angle of refraction is 90°.
Optical fibre
A thin glass fibre that carries light by total internal reflection.
Focal length
The distance from the lens to its principal focus.
Power of a lens
The reciprocal of the focal length in metres, measured in dioptres.
Accommodation
The ability of the eye to change its focal length to focus on near or far objects.
Near point
The closest distance at which the eye can see clearly, about 25 cm.
Angular magnification
The ratio of the angle subtended by the image to that by the object at the unaided eye.
Normal adjustment
A telescope setting with the final image at infinity.
Formula sheet
| Name | Formula | What it means |
|---|---|---|
| Refractive index | n = c/v | Speed of light in vacuum over speed in the medium. |
| Snell's law | n₁ sin θ₁ = n₂ sin θ₂ | Angles are measured from the normal. |
| Critical angle | sin C = 1/n | From the denser medium into air. |
| Lens formula | 1/f = 1/p + 1/q | Real is positive. |
| Magnification of a lens | m = −q/p | Negative means inverted. |
| Power of a lens | P = 1/f | f in metres, P in dioptres. |
| Simple microscope | M = 1 + d/f | Image at the near point, d = 25 cm. |
| Telescope | M = f_o/f_e | Normal adjustment. |
Short questions with model answers
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
State Snell's law. (2 marks)
For light passing from one medium to another, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, n₁ sin θ₁ = n₂ sin θ₂, where n₁ and n₂ are the refractive indices of the two media.
What are the conditions for total internal reflection? (2 marks)
Light must travel from a denser medium to a rarer medium, and the angle of incidence in the denser medium must be greater than the critical angle.
How does an optical fibre guide light? State two uses. (3 marks)
An optical fibre has a glass core of high refractive index surrounded by a cladding of lower refractive index. Light enters the core at a small angle to the axis, and strikes the boundary at an angle greater than the critical angle, so it is totally internally reflected again and again along the fibre with very little loss. Uses are telecommunication and internet cables, and the endoscope in medicine.
Name the defects of vision and the lenses used to correct short sight and long sight. (3 marks)
Short sight (myopia) makes distant objects blurred because the image forms in front of the retina. A concave lens corrects it. Long sight (hypermetropia) makes near objects blurred because the image forms behind the retina. A convex lens corrects it. Astigmatism and presbyopia are also common defects.
Explain the working of a compound microscope with a ray diagram described in words. (5 marks)
A compound microscope has an objective of short focal length and an eyepiece of longer focal length, placed on the same axis. The small object is put just outside the focal point of the objective. The objective forms a real, inverted and magnified image inside the tube. This image lies within the focal length of the eyepiece, which acts as a simple magnifier and forms a final virtual, inverted and further magnified image, usually at the near point (25 cm) of the eye. The total magnification is the product of the magnification of the objective and that of the eyepiece, M = M_o × (1 + d/f_e). Using lenses of very short focal lengths gives a high magnification.
Why is the objective of an astronomical telescope made with a large aperture? (2 marks)
A large aperture collects more light from the faint distant object, making the image brighter, and it also improves the resolving power, so close objects can be separated.
Numericals with worked steps
Find the critical angle for glass of refractive index 1.5.
- sin C = 1/n = 1/1.5 = 0.667.
- C = sin⁻¹(0.667) = 41.8°.
Answer: 41.8°.
An object is 30 cm from a convex lens of focal length 10 cm. Find the image distance and the magnification.
- 1/q = 1/f − 1/p = 1/10 − 1/30 = 3/30 − 1/30 = 2/30.
- q = 15 cm (real image on the other side).
- m = −q/p = −15/30 = −0.5.
Answer: q = 15 cm, m = −0.5 (real, inverted and half the size).
A magnifying glass has a focal length of 5 cm. Find its magnifying power when the image is at the near point and when it is at infinity. Take d = 25 cm.
- Image at the near point: M = 1 + d/f = 1 + 25/5 = 6.
- Image at infinity: M = d/f = 25/5 = 5.
Answer: 6 and 5.
An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. Find the magnifying power and the length of the telescope in normal adjustment.
- M = f_o/f_e = 100/5 = 20.
- Length = f_o + f_e = 100 + 5 = 105 cm.
Answer: M = 20 and length = 105 cm.
A short sighted person has a far point at 50 cm. Find the power of the lens needed to see distant objects clearly.
- The lens must form an image of a distant object at the far point, so the virtual image distance is 50 cm on the same side and f = −50 cm = −0.5 m.
- P = 1/f = 1/(−0.5) = −2.
Answer: −2 D (a concave lens).
Common mistakes
- Measuring angles from the surface instead of from the normal in Snell's law.
- Using the critical angle formula sin C = 1/n for light travelling from a rarer to a denser medium.
- Forgetting to convert the focal length to metres when calculating power in dioptres.
- Using the wrong sign convention in the lens formula and mixing real and virtual images.
- Confusing the lens of a myopic eye (concave correction) with that of a hypermetropic eye (convex correction).