Step 1: Understanding the Concept:
The power of a lens is a measure of its ability to converge or diverge light rays. It is defined as the reciprocal of the focal length of the lens. The sign convention for focal length is crucial.
Step 2: Key Formula or Approach:
The power \(P\) of a lens is given by:
\[ P = \frac{1}{f} \]
where the focal length \(f\) must be expressed in meters for the power \(P\) to be in dioptres (D).
By sign convention, a concave lens (diverging lens) has a negative focal length.
Step 3: Detailed Explanation:
Given data:
Lens type: Concave lens.
Focal length magnitude = 50 cm.
Step 1: Apply sign convention.
For a concave lens, the focal length is negative.
So, \(f = -50 \, \text{cm}\).
Step 2: Convert focal length to meters.
\[ f = -50 \, \text{cm} = -0.5 \, \text{m} \]
Step 3: Calculate the power.
\[ P = \frac{1}{f} = \frac{1}{-0.5 \, \text{m}} = -2 \, \text{D} \]
Step 4: Final Answer:
The power of the concave lens is -2 D.
Match List-I with List-II
\[\begin{array}{|l|l|} \hline \text{List-I (Soil component)} & \text{List-II (Definition)} \\ \hline (A)~\text{Azonal soils} & (I)~\text{An individual natural aggregate of soil particles.} \\ (B)~\text{Regoliths} & (II)~\text{Organisms living in the soil or ground} \\ (C)~\text{Ped} & (III)~\text{Soils have uniformity from the top-surface to the base, and do not have well-developed soil horizons.} \\ (D)~\text{Edaphons} & (IV)~\text{Zone of loose and unconsolidated weathered rock materials.} \\ \hline \end{array}\]
Choose the correct answer from the options given below:
Match List-I with List-II
\[\begin{array}{|l|l|} \hline \text{List I Content of humus} & \text{List II Percentage of contents} \\ \hline \text{(A) Carbon} & \text{(I) 35-40\%} \\ \hline \text{(B) Oxygen} & \text{(II) ~5\%} \\ \hline \text{(C) Hydrogen} & \text{(III) 55-60\%} \\ \hline \text{(D) Nitrogen} & \text{(IV) 15\%} \\ \hline \end{array}\]
Choose the correct answer from the options given below: