Geometrical Optics flash cards

1. 1.State the two laws of reflection.

   (1) The incident ray, reflected ray and normal all lie in the same plane. (2) The angle of incidence equals the angle of reflection: $i=r$ (measured from the normal).

   Hint: Two laws — plane and angles.

2. 2.By what angle does a reflected ray rotate if the mirror is rotated by angle $\theta$ (incident ray fixed)?

   The reflected ray turns by $2\theta$. This is the basis of the mirror-galvanometer (lamp-and-scale) arrangement.

   Hint: Mirror turns $\theta$, ray turns double.

3. 3.How many images are formed by two plane mirrors inclined at angle $\theta$?

   If $360/\theta$ is even: $n=\frac{360}{\theta}-1$. If odd and object off bisector: $n=\frac{360}{\theta}-1$; if on bisector: $n=\frac{360}{\theta}$. If $360/\theta$ is not an integer, take the integer part of $360/\theta$.

   Hint: $n=\frac{360}{\theta}-1$ with parity rules.

4. 4.How far does a plane mirror image move if the mirror moves toward/away from a fixed object?

   When the mirror moves a distance $d$ (object fixed), the image moves $2d$. Image speed relative to object is twice the mirror's speed.

   Hint: Image shifts double the mirror shift.

5. 5.A person of height $H$ needs what minimum vertical plane-mirror length to see their full image?

   Minimum length $=H/2$, mounted so its top is midway between eye-level and head-top. Result is independent of the distance from the mirror.

   Hint: Half your height, distance-independent.

6. 6.What is the nature and characteristics of a plane-mirror image?

   Virtual, erect, same size as object, laterally inverted, and located as far behind the mirror as the object is in front.

   Hint: Virtual, erect, same size, laterally inverted.

7. 7.Define centre of curvature, pole and radius of curvature of a spherical mirror.

   Centre of curvature $C$ = centre of the sphere the mirror is a part of. Pole $P$ = geometric centre of the mirror surface. Radius of curvature $R$ = distance $PC$.

   Hint: $C$, $P$, and $R=PC$.

8. 8.Relation between focal length and radius of curvature for a spherical mirror.

   $f=\dfrac{R}{2}$. The focus lies midway between pole and centre of curvature (valid for paraxial rays).

   Hint: Focal length is half of $R$.

9. 9.State the mirror formula and the sign convention used.

   $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$. New Cartesian convention: distances measured from the pole, along incident-light direction positive; heights up positive. For concave mirror $f<0$, convex $f>0$.

   Hint: $1/v+1/u=1/f$; measure from pole.

10. 10.Define linear (transverse) magnification for a mirror and its sign meaning.

    $m=\dfrac{h'}{h}=-\dfrac{v}{u}$. $m>0$: erect virtual image; $m<0$: inverted real image; $|m|>1$ enlarged, $|m|<1$ diminished.

    Hint: $m=-v/u$; sign tells erect/inverted.

11. 11.Give the longitudinal magnification of a small object along the mirror axis.

    For a short object along the axis, $m_L=-\dfrac{dv}{du}=-\dfrac{v^2}{u^2}=-m^2$ (transverse $m=-v/u$). Longitudinal magnification $=$ (transverse magnification)$^2$ in size.

    Hint: It goes as $m^2$.

12. 12.Where must an object be placed before a concave mirror to get a real, inverted, magnified image?

    Between focus $F$ and centre of curvature $C$ ($f<|u|<2f$). Image forms beyond $C$, real, inverted, magnified.

    Hint: Object between $F$ and $C$.

13. 13.For a concave mirror, when is the image virtual and erect?

    When the object lies between pole and focus ($|u|<f$). Image is virtual, erect and magnified, formed behind the mirror.

    Hint: Object inside the focus.

14. 14.Describe images formed by a convex mirror for all object positions.

    Always virtual, erect, diminished, and located between pole and focus behind the mirror. This wide field of view is why convex mirrors are used as rear-view/vehicle mirrors.

    Hint: Always virtual, erect, diminished.

15. 15.State Snell's law of refraction.

    $n_1\sin i=n_2\sin r$, i.e. $\dfrac{\sin i}{\sin r}=\dfrac{n_2}{n_1}=n_{21}$ constant. Incident ray, refracted ray and normal are coplanar.

    Hint: $n_1\sin i=n_2\sin r$.

16. 16.Define absolute refractive index and its relation to wave speed.

    $n=\dfrac{c}{v}$, ratio of speed of light in vacuum to that in the medium. Since $c=n_1 v_1=n_2 v_2$, denser media (higher $n$) have slower light.

    Hint: $n=c/v$.

17. 17.What happens to frequency, wavelength and speed of light on entering a denser medium?

    Frequency stays constant. Speed decreases: $v=c/n$. Wavelength decreases: $\lambda_{med}=\lambda_{vac}/n$. Colour (frequency) is unchanged.

    Hint: Frequency fixed; $\lambda$ and $v$ shrink.

18. 18.State the principle of reversibility of light and one consequence.

    A light ray retraces its path if its direction is reversed. Consequence: $n_{12}\times n_{21}=1$, i.e. $n_{21}=1/n_{12}$.

    Hint: Path retraces; $n_{12}n_{21}=1$.

19. 19.How much does a glass slab of thickness $t$ and index $n$ shift a ray laterally (near-normal incidence)?

    Lateral shift $d=t\Big(1-\dfrac{1}{n}\Big)$ for small angles; general $d=\dfrac{t\sin(i-r)}{\cos r}$. The emergent ray is parallel to the incident ray but displaced.

    Hint: Parallel emergent ray, shifted.

20. 20.Apparent (normal) shift for an object viewed through a slab / from denser medium.

    Normal shift (object appears raised) $=t\Big(1-\dfrac{1}{n}\Big)$. Apparent depth $=\dfrac{\text{real depth}}{n}$ when viewed from air above water.

    Hint: Apparent depth $=$ real depth$/n$.

21. 21.A pool of real depth $h$ is viewed normally from air. What is its apparent depth?

    $h_{app}=\dfrac{h}{n_{water}}$. With $n=4/3$, apparent depth $=\dfrac{3h}{4}$, so the bottom looks raised.

    Hint: Divide by refractive index.

22. 22.State the refraction formula at a single spherical surface.

    $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$, with distances measured from the pole and $R$ = radius of curvature of the surface.

    Hint: $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$.

23. 23.Transverse magnification for refraction at a single spherical surface.

    $m=\dfrac{h'}{h}=\dfrac{n_1 v}{n_2 u}$. This reduces to $-v/u$ when both media are the same.

    Hint: $m=\frac{n_1 v}{n_2 u}$.

Open the interactive deck for the other 68 cards, with self-grading so the ones you keep missing come back.