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Question

The defect of hypermetropia can be corrected by using which of the following?

The correct answer is
Convex lens

Understanding Hypermetropia

Hypermetropia, also known as farsightedness, is a vision defect where a person can see distant objects clearly but faces difficulty seeing near objects distinctly. This happens because the eye's optical system converges light rays to a focal point behind the retina, rather than directly on it.

Correcting Hypermetropia

To correct hypermetropia, a lens is needed that increases the overall converging power of the eye. This helps to bend the light rays more sharply so that the image is focused precisely on the retina.

  • Convex Lens: A convex lens is a converging lens. When placed in front of the eye, it adds to the eye's refractive power, causing light rays to converge sooner and focus on the retina. This effectively corrects the defect of hypermetropia.
  • Concave Lens: Concave lenses are diverging lenses and are used to correct myopia (nearsightedness), where the focal point is in front of the retina.
  • Cylindrical Lens: These lenses are primarily used to correct astigmatism, a condition where the eye's refractive power is different in different meridians.
  • Prism: Prisms are used to correct conditions involving eye alignment or double vision, not refractive errors like hypermetropia.

Therefore, a convex lens is the appropriate choice for correcting hypermetropia.

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Important Questions from Refraction and Reflection

  1. A convex lens of focal length f will form a magnified real image of an object, if the object is placed.

  2. A ray of light travelling in the direction \(\frac{1}{2} (\hat i + \sqrt 3 \hat j)\) is incident on a plane mirror. After reflection it travels along the direction  \(\frac{1}{2} (\hat i - \sqrt 3 \hat j)\)  The angle of incidence is:

  3. Twinkling of stars is due to atmospheric

  4. An optical fibre has a core material of refractive index of 1.55 and cladding material of refractive index of 1.50. The numerical aperture of the fibre is

  5. The refracting angle of a prism is $A$. The refractive index of the material of the prism is $\frac{1+\cos A}{\sin A}$. The angle of minimum deviation is:
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