An Object Sits at a Distance of F/2 From a Convex Lens, Where F Is the Focal Length.

Pure mathematics optics

Snell's law

Problem:

A plane wave is propagating from exclusive material 1 with index of refraction n₁ toward a plane user interface with material 2, as shown.  The index of refraction of corporate 2 is n₂ = 1.  For angles θ_i > 50^o the wave is totally echolike.
(a)  Find the index of refraction n₁ of bodied 1.
(b)  For θ_i = 0 find the transmission T.

Solution:

• Concepts:
  Snell's law: n₁sinθ_i = n₂sinθ_t, total internal contemplation, transmission coefficient and transmittance
• Rational:
  If n₂ < n₁, then for angles θ_i > sin^-1(n₂/n₁) we get total internal observation.
  The transmittal coefficient for normal incidence is t_12s = t_12p = 2n₁/(n₂ + n₁) if we assume μ₁ = μ₂ = μ₀.
• Details of the calculation:
  (a)  sinning(50^o) = 1/n₁.  n₁ = 1.3.
  (b)  t_12s = t_12p = 2n₁/(n₂ + n₁) = 1.13.
  T = <S _t>∙n/<S _i>∙n where n is a unit vector normal to the interface.
  T = (n₂/n₁)|t₁₂|² = 0.986

Problem:

A beam of colour vision deficiency light traveling finished air strikes the top of a flat slab of glass at an angle 60^o to the regular, passes through the glass, and emerges from the undersurface of the slab.  The ice has a thickness t and a deflective index n = 1.52.

(a)  What is the angle of deflexion where the light enters the glass?
(b)  Show that the transmit emerging from the glass is parallel to the incident beam.

Solution:

• Concepts:
  Snell's law
• Reasoning:
  We apply Snell's police force at both interfaces.
• Details of the calculation:
  (a)  n₁ sinθ₁ = n₂ sinθ₂,
  sinθ₂ = (n₁/n₂) sinθ₁ = (1/1.52) sin 60^o = 0.569
  θ₂ = 34.7^o
  (b)  Substitute from component (a) and use n₁ = n₃.
  n₂ sinθ₂ = n₃ sinθ₃,
  n₂ (n₁/n₂) sinθ₁ = n₁ sinθ₃,  sinθ₁ = sinθ₃,  θ₁ = θ₃

Problem:

A dinky origin of light is mounted interior a cylindrical container of height h.  The behind of the container is covered with a mirror.  Initially, the container is white.  And so a legible liquid with the index of refraction n is tardily poured into the container.  The level off of liquid rises steadily, reaching the top of the container one of these days t.Uncovering the speed of the image of the source during this outgrowth.

Solution:

• Concepts:
  Expression and refraction
• Reasoning:
  The light is refracted at at the bound between thawed and air and mirrored at the bottom of the container.
  Ignore reflections from the surface of the liquid.
• Inside information of the calculation:
  Let the source personify located a distance d above the bottom of the container, d < h.
  The height of the disposable at time t' is h' = (h/t)t'.
  Trace two rays departure the source, one vertical, and one devising a small angle θ with the passant.  Their (virtual) intersection Simon Marks the image position
  (a)  Take for granted h' < d.
  
  Applying the law of refraction sinθ = nsinθ' in the small angle idea we have θ = nθ'.
  Trigonometry then yields (d - h')θ + 2h'θ/n = (x + h')θ operating room x = d - 2h' + 2h'/n.
  Here x is the distance of the image below the bottom of the container.
  dx/dt' = -2h/t + 2h/(Tennessee) = -(2h/t)(1 - 1/n).
  So x is decreasing at a steady rate, the image of the origin is soaring up with speed
  v = (2h/t)(1 - 1/n).

  (b)  Assume h' > d < h.
  
  In the small Angle approximation trigonometry yields
  dθ/n + h'θ/n = (x + h')θ.
  Notation: x is negative if the image position is above the bottom of the container.
  x = d/n + h'/n - h'.  dx/dt = h/(nt) - h/t = -(h/t)(1 - 1/n)
  Thusly x is still decreasing, the image of the source is now haunting up with speed
  v = (h/t)(1 - 1/n).

Job:

A parallel beam of monochromatic illuminated strikes a transparent prism.  The Cross section of the prism is a uniform hexagon.  The beam is comparable to the "top" and "bottom" faces of the prism, and points A and B in the diagram are the midpoints of the corresponding edges.

After the refraction, two separate parallel beams of light come out from the optical prism. What is the minimum index of refraction of the stuff of the prism that allows such an effect?

Solution:

• Concepts:
  Snell's law
• Intelligent:
  Debate the diagram of the hexagonal prism below, along with the part of the light beam entering 'tween points C and A

  

  Since the side containing A and C is parallel to the side containing A' and C', the beam emerging from the strip A'C' bequeath be parallel to the beam entering the strip AC.  By symmetry, another collimate beam exits from the face just above point A'.  The diagram is for the minimum index of refraction needed to have two parallel beams.  In the diagram the separation 'tween the beams is nix.  A larger indicant moves points A' and C' down their side, creating a "blank" region 'tween the two nonintersecting beams.

• Details of the computing:
  Think that the hexagon has side length L.  The distance from midpoint A to centre C' is
  2L cos30^o = (√3)L.  This segment also serves as the normal line to the rays entering Oregon exiting the prism at A or C'.  The length of the ray impermanent from A to A' can be found victimisation the Philosopher theorem,  ((L/2)² + 3L²)^½ = (√13)L/2.
  Snell's law:  sin30^o = n sinθ = n/(√13),  n = (√13)/2 = 1.803.

---

Lenses and mirrors

Problem:

A thin lens creates a sharp project of the object onto the wall.  The aloofness between the targe and the wall is ninefold the distance from the wall to the closest focal point. What is the object magnification?

Solution:

• Concepts:
  The thin lens
• Reasoning:
  We can role the lens equation, 1/x_o + 1/x_i = 1/f, and M = -x_i/x_o to solve the problem.
• Details of the calculation:
  Given:  A real image implies a convex lens, the visualise is inverted, f is positive, M is negative.
  
  x_i + x_o = 9(XI - f),  x_o = 8x_i - 9f.
  1/(8x_i - 9f) + 1/x_i = 1/f --> f² - 2x_if + (8/9)x_i ² = 0,  f = x_i - (x_i ² - (8/9)x_i ²)^½,
  since f < x_i.
  f = (2/3)x_i,   x_i = (3/2)f,   x_o = 3f,   M = -x_i/x_o = -½.

Problem:

Say a certain targe with a height of 1 cm is located 10 cm in front of its figure of speech, which has a summit of 0.5 cm.  Some object and trope are upright.  Find the location and focal distance of a single shrivelled lens that performs this imaging.

Solution:

• Concepts:
  The thin genus Lens equation
• Reasoning:
  This is a fiddle-shaped covering of the thin lens equation.
• Inside information of the calculation:
  The image formed by the lens is rearing.  It therefore is a virtual image.  The visualize is smaller than the object, it therefore is spoon-shaped away a diverging lens.
  M = -x_i/x_o = 0.5,  x_i = -0.5 x_o, image and object are on the synoptic side of the lens.
  x_o + x_i = 10 centimeter,  x_o - 0.5 x_o = 10 curium,  x_o = 20 cm
  1/x_o + 1/x_i = 1/f,  1/20 - 1/10 = 1/f,  f = -20 Cm

Problem:

An targe is placed 8 cm from a thin double convex lens of focal length 12 cm?
(a)  Find the image position and ascertain whether the image is real Beaver State realistic by victimisation the lens convention.
(b)  Repeat part (a) by using visual communication construction instead of the crystalline lens expression.

Solution:

• Concepts:
  The thin crystalline lens
• Reasoning:
  We are theoretic to use the lens equation and a graphical construction to regain the image position for a thin lens.
• Details of the calculation:
  (a)  Convex lens --> f is positive.
  1/x_o + 1/x_i = 1/f.  1/x_i = 1/12 - (1/8) = -1/24.  x_i = - 24, we have a virtual image.
  (b)  Graphical construction:

  

Job:

A concave spherical mirror is used by a dentist to produce an enlarged image of a tooth.  If the radius of curve of the mirror is 2.0 cm, how close is the mirror to the tooth when the paradigm appears triple the size of the tooth?  Is the image erect OR inverted?

Solution:

• Concepts:
  Spherical mirrors
• Reasoning:
  If an object is placed before of a spherical mirror at an object distance x_o, then an image is formed at an pictur distance x_i, where x_o and x_i satisfy the mirror par, 1/x_o + 1/x_i = 1/f.
  House conventions for the mirror equivalence:
  x_i is positive for a real image in front of the mirror aerofoil, and x_i is negative for a virtual image behind the mirror surface.  x_o and x_i are the perpendicular distances from the centrist aeroplane of the mirror as shown in the drawing along the right.
  The focal distance f and the radius of curvature R = 2f are positive for a concave mirror and negative for a convex mirror.
  The enlargement is M = h_i/h_o= -x_i/x_o.  If the magnification is perverse, the prototype is inverted.
• Inside information of the calculation:
  The effigy is a virtual upright look-alike.  f = 1 Cm.  M = -3, x_i = -3x_o.  1/x_o  - (1/3)x_o = 2/(3x_o) = 1/(1 cm).
  x_o = (2/3) Cm.

Job:

Consider optical components central on the x-axis.
A thin converging lens with point length f = 2 cm is located at x = 0, and a convex mirror with radius of curvature |R| = 4 cm is located at x = d.
An object is placed at x = -3 centimetre.  Find the pictur location and magnification if
(a) d = 10 cm,
(b) d = 5 cm.
Is the image upright Beaver State inverted?

Solution:

• Concepts:
  Mirrors and lenses
• Reasoning:
  The image formed by the lens becomes the object for the mirror, the image formed by the mirror becomes the physical object for the lens after reflection.
• Details of the calculation:
  (a)  1.  Lens:  1/x_o + 1/x_i = 1/f, (1/3) + 1/x_i = ½, x_i = 6.  The image is located at x = 6 cm, 6 Cm behind the lens system and 4 cm in front of the mirror.
  The magnification is M₁ = -6/3 = -2.
  2.  Mirror:  f = -2,  ¼ + 1/x_i = -½, x_i = -4/3.  The image is located at x = 10 plus 4/3 atomic number 96, 4/3 Cm behind the mirror, and 34/3 cm to the right of the crystalline lens.
  M₂ = (1/3).
  3.  3/34 + 1/x_i = ½.  x_i = 17/7.  The image is located at x = -17/7 cm.
  M₃ = -3/14.
  The magnification of the system is M₁*M₂*M₃ = 1/7.  The image is perpendicular.
  (b)  1.  Lens:  1/x_o + 1/x_i = 1/f, (1/3) + 1/x_i = ½, x_i = 6.  The image is located at x = 6 cm, 6 cm behind the lens and 1 centimetre in behind the mirror.
  The magnification is M₁ = -6/3 = -2.
  2.  Mirror:  f = -2,  the object distance is pessimistic.
  -1/1 + 1/x_i = -½, x_i = 2.  The image is located at x = 3 cm, 2 Cm before the mirror, and 3 cm to the right hand of the lens.
  M₂ = 2.
  3.  (1/3) + 1/x_i = ½.  x_i = 6.  The image is located at x = -6 cm.
  M₃ = -6/3 = -2.
  The magnification of the system is M₁*M₂*M₃ = 8.  The image is upright.

Problem:

An optical system comprises in plough, from left to right: an perceiver, a lens of focal length +30 centimeter, an erect object 20 mm high, and a convex mirror of radius 80 centimeter. The object is between the lense and the mirror, 20 cm from the lens and 50 cm from the mirror. The observer views the prototype that is formed first by reflection from the convex mirror and then by refraction from the lens. What is the position of this final image, measured from the mirror, and what is its elevation?

Solution:

• Concepts:
  Mirrors and lenses, 1/x_o + 1/x_i = 1/f
• Logical thinking:
  The image s-shaped by the mirror becomes the physical object for the electron lens after reflection.
• Inside information of the calculation:
  reflection:  f = -|R|/2 = - 40 cm.  1/50 + 1/x_i = -1/40,  x_i = -200/9 = 22.22.
  The intermediate image is 200/9 cm to the mighty (behind) the mirror.
  M₁ = -x_i/x_o = 4/9 = 0.444.  The ikon is virtual and upright.
  refraction:  x_o = 830/9 cm.  9/830 +  1/x_i = 1/30,  x_i = 44.46.
  The final image is 44.46 cm to the left (in movement of) the lens.
  It is 114.46 cm to the left of the mirror.
  M₂ = -x_i/x_o = -44.46/92.22 = -0.482.  The image is real and inverted.
  M_final = M₁M₂ = -0.214.  The height of the unalterable epitome is 4.3 mm and it is real and inverted.
  

An Object Sits at a Distance of F/2 From a Convex Lens, Where F Is the Focal Length.

Source: http://electron6.phys.utk.edu/PhysicsProblems/E&M/7-EM%20waves/Geometrical.html

Share this post