Human Eye

Exemplar Problems

MCQs Part 1

Question 1: A person cannot see distinctly objects kept beyond 2 m. This defect can be corrected by using a lens of power

  1. + 0.5 D
  2. – 0.5 D
  3. + 0.2 D
  4. – 0.2 D

Answer: (b) – 0.5 D

Explanation: The far point for this person is 2 m, while the far point of a person with normal vision is 1 m. If image of the object kept at 2 m is formed at 1 m then the person shall be able to clearly see that object.

Object distance u = -2 m and image distance v = -1 m

Focal length of required lens can be calculated by using lens formula

`1/v-1/u=1/f`

Or, `1/f=-1+1/2=-1/2`

Or, `f=-2` m

Now, power `D=1/f`

Or, `D=-1/2=-0.5` D


Question 2: A student sitting on the last bench can read the letter written on the blackboard but is not able to read the letters written in his textbook. Which of the following statements is correct?

  1. The near point of his eyes has receded away
  2. The near point of his eyes has come closer to him
  3. The far point of his eyes has come closer to him
  4. The far point of his eyes has receded away

Answer: (a) The near point of his eyes has receded away

Explanation: The student is suffering from hypermetropia and hence is unable to see nearby object. So, the near point of his eyes has receded away

Question 3: A prism ABC (with BC as base) is placed in different orientations. A narrow beam of white light is incident on it as shown in figure. In which of the following cases, after dispersion, the third colour from the top corresponds to the colour of the sky?

  1. (i)
  2. (ii)
  3. (iii)
  4. (iv)
Prisms with incident light Prisms with incident light

Answer: (b) (ii)

Explanation: When a prism kept with it base at the bottom, the seven colours of the rainbow appear from bottom to top in the sequence VIBGYOR, i.e. blue (colour of sky) is third from bottom. Figure 2 is showing base at the top. So, third colour from top will be blue.


Question 4: At noon the sun appears white as

  1. Light is least scattered
  2. All the colours of the white light are scattered away
  3. Blue colour is scattered the most
  4. Red colour is scattered the most

Answer: (a) Light is least scattered

Explanation: As light is least scattered at noon, so all the seven colours of visible spectrum reach our eyes. So, the sun appears white at noon.

Question 5: Which of the following phenomena of light are involved in the formation of a rainbow?

  1. Reflection, refraction and dispersion
  2. Refraction, dispersion and total internal reflection
  3. Refraction, dispersion and internal reflection
  4. Dispersion, scattering and total internal reflection

Answer: (c) Refraction, dispersion and internal reflection

Explanation: Water droplets work like prism. They refract and disperse the incident light. Refraction bends the light and dispersion causes segregation of all the colours of white light. After that they reflect the light so that rainbow is formed on the opposite side of sun in the sky.


Question 6: Twinkling of stars is due to atmospheric

  1. Dispersion of light by water droplets
  2. Refraction of light by different layers of varying refractive indices
  3. Scattering of light by dust particles
  4. Internal reflection of light by clouds

Answer: (b) Refraction of light by different layers of varying refractive indices

Explanation: Due to refraction of starlight through different layers of varying refractive indices, the apparent position of the source of light keeps on changing. Due to this, stars appear to twinkle.

Question 7: The clear sky appears blue because

  1. Blue light gets absorbed in the atmosphere
  2. Ultraviolet radiations are absorbed in the atmosphere
  3. Violet and blue lights get scattered more than lights of all other colours by the atmosphere
  4. Light of all other colours is scattered more than the violet and blue colour lights by the atmosphere

Answer: (c) Violet and blue lights get scattered more than lights of all other colours by the atmosphere

Explanation: The molecules of air and fine particles in air are smaller than the wavelength of visible spectrum. These particles are more effective at scattering the light of shorter wavelength. So, they scatter light towards the blue end of the spectrum more strongly than the light towards the red spectrum.



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