Optics

30 flashcards · Other · by inquisitive learner

1. Gauss's Law for E

   ∇\nabla∇ ⋅\cdot⋅ E⃗\vec{E}E === ρϵ0\frac{\rho}{\epsilon_0}ϵ0​ρ​

2. diffraction

   when a wave passes through a barrier that has an aperture or edge

3. Huygens-Fresnel principle

   view every point on a wavefront as a source of spherical wavelets, and the secondary wavelets emanating from different points mutually interfere

4. Fresnel diffraction (near-field)

   diffraction that occurs when the both the point source and the screen are relatively close to the object forming the diffraction pattern

5. Fraunhofer diffraction (far-field)

   source, obstacle and screen are far enough apart that all lines from the source to the obstacle can be considered to be parallel

6. dark fringes, single slit diffraction

   sin⁡θ\sin{\theta}sinθ === mλa\frac{m\lambda}{a}amλ​ (m=±1,±2,±3,...)(m = \pm 1, \pm 2, \pm 3, ...)(m=±1,±2,±3,...) sin⁡θ=\sin{\theta} =sinθ= angle of line from the centre of t slit to the mth dark fringe on the screen m=m =m= mth dark fringe λ=\lambda =λ= wavelength a=a =a= slit width

7. sin⁡θ=0\sin{\theta}=0sinθ=0 corresponds to a bright band

8. b^2/\lambda"> R>b2/λR > b^2/\lambdaR>b2/λ is Fraunhofer| far-field diffraction

9. Fresnel diffraction is R < b^2/lambda

10. If the screen is distant, the rays to the screen are approximately parallel

11. Rayleigh criterion

    sin⁡θ1=1.22λD\sin{\theta_1} = 1.22 \frac{\lambda}{D}sinθ1​=1.22Dλ​

12. conditions for constructive interference

    path lengths differ by integer wavelengths r2−r1=r_2 - r_1 =r2​−r1​= mλm\lambdamλ ​

13. conditions for destructive interference

    path lengths differ by half-integer wavelengths r2−r1=r_2-r_1 =r2​−r1​= ((( m+12m+ \frac{1}{2}m+21​ ))) λ\lambdaλ

14. Intensity of an electromagnetic wave

    I=I =I= ε0\varepsilon_0ε0​ ccc ⟨\lang⟨ ∣|∣ E2⃗\vec{E^2}E2 ∣|∣ ⟩T\rang_T⟩T​

15. Through a double slit, bright bands are the result of wavefronts arriving in phase and interfering constructively

16. Bright fringes, double slit

    ymy_mym​ === RRR mλd\frac{m\lambda}{d}dmλ​ ym=y_m =ym​= bright fringe - central fringe distance R=R = R= slit-screen distance λ=\lambda =λ= wavelength d=d = d= inter-slit distance

17. path difference, double slit

    r2−r1r_2 - r_1r2​−r1​ =dsin⁡θ= d\sin{\theta}=dsinθ

18. image of a point object: spherical refracting surface

    nas\frac{n_a}{s}sna​​ +++ nbs′\frac{n_b}{s'}s′nb​​ === (nb−na)R\frac{(n_b - n_a)}{R}R(nb​−na​)​ na=n_a =na​= refractive index of first medium nb=n_b =nb​= refractive index of the second medium s=s =s= object distance s′=s' =s′= image distance R=R =R= radius of curvature

19. When is the object distance positive?

    When the incoming light and object are on the same side of the reflecting or refracting surface

20. Image distance is negative when the image is not on the same side as the outgoing light

21. Magnification Law for Refracting Surfaces

    m=m =m= y′y\frac{y'}{y}yy′​ === −-− nas′nbs\frac{n_a s'}{n_b s}nb​sna​s′​ m=m =m= magnification y′=y' =y′= image height y=y=y= object height

22. In the special case of a plane refracting surfaces, the radius of curvature can be taken to be infinite and the magnification is 1

23. lens

    optical system with two refracting surfaces

24. The focal length for a converging thin lens is positive

25. In a thin lens, the focal lengths are equal even when the two sides have different curvatures

26. What sign does a diverging lens' focal length have?

    The focal length of a diverging lens is negative

27. Lensmaker's equation

    1f\frac{1}{f}f1​ === (n−1)(n-1)(n−1) (1R1−1R2)(\frac{1}{R_1} - \frac{1}{R_2})(R1​1​−R2​1​) f=f = f= focal length n=n = n= lens' refractive index R1R_1R1​ === radius of curvature of first surface R2R_2R2​ === radius of curvature of second surface

28. Law of Reflection

    θa\theta_aθa​ === θr\theta_rθr​

29. Snell's Law (Law of refraction)

    nasin⁡θan_a \sin{\theta_a}na​sinθa​ === nbsin⁡θbn_b \sin{\theta_b}nb​sinθb​

30. critical angle

    sin⁡θc\sin{\theta_c}sinθc​ === nbna\frac{n_b}{n_a}na​nb​​