Two convex lenses have focal lengths of 50 cm and 25 cm, respectively. If these two lenses are placed in contact, then the net power of this combination will be equal to
+6 dioptre
This question asks us to find the net power of a combination of two convex lenses placed in contact. To solve this, we first need to understand the concept of the power of a lens and how it relates to focal length, and then apply the rule for combining powers when lenses are in contact.
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 \(f\). The standard unit for lens power is the dioptre (D), which is defined as the power of a lens with a focal length of 1 meter.
The formula for lens power \(P\) is:
\[P = \frac{1}{f}\]Where \(f\) is the focal length in meters.
For convex lenses, the focal length is considered positive, resulting in positive power. For concave lenses, the focal length is negative, resulting in negative power.
We are given the focal lengths of two convex lenses:
To calculate the power in dioptres, we must convert the focal lengths from centimeters to meters:
Now, we can calculate the power of each lens using the formula \(P = 1/f\):
Since both lenses are convex, their powers are positive, which is consistent with our calculations.
When two thin lenses with powers \(P_1\) and \(P_2\) are placed in contact, the net power of the combination \(P_{net}\) is simply the sum of their individual powers. This additive property makes calculating the combined power straightforward.
The formula for the net power of lenses in contact is:
\[P_{net} = P_1 + P_2\]Using the individual powers we calculated:
\[P_{net} = +2 \text{ D} + +4 \text{ D} = +6 \text{ D}\]The net power of the combination of these two convex lenses is +6 dioptres.
The calculated net power of the combination is +6 dioptres. We compare this result with the given options:
Our calculated net power (+6 D) matches Option 2.
| Lens | Focal Length (cm) | Focal Length (m) | Power (Dioptres) |
|---|---|---|---|
| Lens 1 (Convex) | 50 | 0.50 | \(P_1 = 1/0.50 = +2\) |
| Lens 2 (Convex) | 25 | 0.25 | \(P_2 = 1/0.25 = +4\) |
| Combination in Contact | - | - | \(P_{net} = P_1 + P_2 = +2 + +4 = +6\) |
| Concept | Formula | Notes |
|---|---|---|
| Lens Power | \(P = \frac{1}{f}\) | \(f\) must be in meters; \(P\) is in dioptres (D) |
| Net Power of Lenses in Contact | \(P_{net} = P_1 + P_2 + P_3 + \dots\) | Applies to any number of thin lenses in contact |
| Equivalent Focal Length (Lenses in Contact) | \(\frac{1}{F_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \dots\) | Where \(F_{eq}\) is the focal length of the combination |
Understanding lens combinations is crucial in designing optical instruments like telescopes and microscopes. The type of lens (convex or concave) significantly affects the power and the nature of the image formed.
In this specific problem, since both are convex lenses and are in contact, their positive powers add up, resulting in a combination with greater positive power, which means a shorter equivalent focal length and stronger converging ability.
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