To find the power of a lens, we use the formula:
\(P = \frac{1}{f}\)
where \(P\) is the power of the lens in diopters (D), and \(f\) is the focal length of the lens in meters.
According to the problem, the focal length \((f)\) is given as 50 cm. First, we need to convert this into meters:
\(f = 50 \, \text{cm} = 0.50 \, \text{m}\)
Since it is a concave lens, the focal length is negative, so:
\(f = -0.50 \, \text{m}\)
Now, substitute the value of \(f\) into the power formula:
\(P = \frac{1}{-0.50}\)
Simplifying this, we get:
\(P = -2 \, \text{D}\)
Thus, the power of the concave lens is -2 D.
Option analysis:
- +5 D: Incorrect, as a positive power indicates a convex lens, not concave.
- -5 D: Incorrect, we calculated the power as -2 D.
- +2 D: Incorrect, positive power is for convex lenses.
- -2 D: Correct, as calculated above.
Thus, the correct answer is -2 D.
Important Questions from Ray Optics and Optical Instruments
The graph correctly representing the variation of image distance v for a convex lens of focal length f versus object distance u is:
Two slits are made 0.1 mm apart, and the screen is placed 2 m away. The fringe separation when a light of wavelength 500 nm is used is:
Resolving power of a telescope can be increased by increasing:
Match List - I with List - II.
List - I List - II (A) Contracting of Eye ball (I) Myopia (B) Controls the shape of eye lens (II) Cornea (C) Elongation of eye ball (III) Ciliary Muscle (D) Control the light entering in eyes (IV) Hypermetropia Choose the correct answer from the options given below:
Four lenses of focal length ±5cm and ±200cm are available for making a telescope. To produce the largest magnification, the focal length of the eyepiece should be:
