11TH GRADE PRACTICE
Physics Problems: Optics, Grade 11
Reflection, refraction, lenses and diffraction.
Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.
Problem statements
The answer options are there to check against, and the walkthrough opens under each problem.
Problem #37
Geometric opticsBeam angle , refractive index of the medium . What is the angle of refraction?
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Answer
Problem #38
Lenses, interference and diffractionDistance from an object to an image of equal size . Determine the focal length and optical power of the lens.
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The image is equal to the object only at d = 2F.
Answer
Problem #41
Geometric opticsHeight of the Sun above the horizon . At what angle to the horizon should a mirror be placed to illuminate the bottom of a vertical well?
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The beam must go vertically downwards, so the angle between the incident and reflected rays is 110°.
Angles with mutually perpendicular sides are equal: the mirror is placed at 55° to the horizon.
Answer
Problem #42
Geometric opticsAt angle of incidence the angle of refraction is . Determine the refractive index, the speed of light in the substance, and the limiting angle of total internal reflection. What will happen at the angle of incidence from matter into air?
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Limiting angle of total reflection.
There is no angle of incidence with 40° refraction: the maximum angle of refraction is 39.3°, and this is achieved at 90° incidence.
Answer
Problem #43
Geometric opticsWhat must be the refractive index of a substance so that the reflected and refracted rays make an angle? ? Express your answer in terms of the angle of incidence .
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The angle of reflection is equal to the angle of incidence, therefore, when the rays are perpendicular, β = 90° − α.
For example, at α = 60° you get n = tan 60° ≈ 1.73.
Answer
Problem #44
Lenses, interference and diffractionThe object is at a distance from the lens, the image in times the size of the item. Determine the possible focal lengths and optical powers for real and virtual images.
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The image is valid.
The image is imaginary - for it f is taken with a minus sign.
Answer
Problem #45
Lenses, interference and diffractionLens with focal length gives an increase in times. Determine the distance between the object and the image. What values are possible if the image is reduced in once?
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Real image: object and image on opposite sides of the lens.
Virtual image: it is on the same side as the object.
Zooming out by 5 times is the same scheme as zooming in, only the object and image are swapped.
Answer
Problem #46
Lenses, interference and diffractionCoherent waves with a frequency interfere in the air . The light will increase or decrease at a point with a path difference ?
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An even number of half-waves means the light will intensify: this is the maximum.
Answer
Problem #47
Lenses, interference and diffractionOn a diffraction grating with a period Monochromatic light falls normally. The third order maximum is visible at an angle . Determine the wavelength as well as the speed and wavelength in glass with . How will the frequency and wavelength change when the frequency is doubled?
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The frequency does not change when passing into glass - the speed and wavelength change.
Doubling the frequency shortens the wave by half - this is ultraviolet.
Answer
Problem #48
Lenses, interference and diffractionThe diffraction grating has cracks on . What is the angle between the second order purple maxima at ?
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The maxima are symmetrical about the center, so the angle between them is twice as large.
Answer
Problem #49
Lenses, interference and diffractionFocal length of the projection lamp lens . The slide is at a distance from the lens. What linear magnification does the flashlight provide?
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The slide is placed a little further from the focus - then the image turns out large and real.
Answer
Problem #52
Lenses, interference and diffractionAt a distance from the screen there is a diffraction grating with a period . At what distance from the central maximum is the second-order violet maximum at ?
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The angles are small, so the sine can be replaced by a tangent.
Answer
The theory behind these problems
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Formulas by topic
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