Sure! Let's solve this step by step.
### Part (a)
The power of a lens (in diopters, D) tells us about the lens' ability to converge or diverge light. The sign of the power determines the nature of the lens.
1. For the lens with power +2D:
- A positive power indicates that the lens is converging. A converging lens is also known as a convex lens.
- Nature of the lens: Converging (Convex) lens.
2. For the lens with power -4D:
- A negative power indicates that the lens is diverging. A diverging lens is also known as a concave lens.
- Nature of the lens: Diverging (Concave) lens.
### Part (b)
Here, we need to find where the object should be placed such that the size of the image is twice the size of the object, given some conditions.
1. Given data:
- The image distance [tex]\( v = -100 \)[/tex] cm (since the image is real and inverted, we take [tex]\( v \)[/tex] as negative).
- Magnification [tex]\( m = -2 \)[/tex] (since the image is real and inverted, magnification is negative, and the size is twice that of the object, so [tex]\( m = -2 \)[/tex]).
2. Finding the object distance [tex]\( u \)[/tex]:
- Magnification [tex]\( m \)[/tex] is given by:
[tex]\[
m = -\frac{v}{u}
\][/tex]
- Rearranging the formula to solve for [tex]\( u \)[/tex]:
[tex]\[
u = -\frac{v}{m}
\][/tex]
- Substituting the values:
[tex]\[
u = -\frac{-100}{-2} = 50 \text{ cm}
\][/tex]
- Therefore, the object should be placed at [tex]\( 50 \)[/tex] cm in front of the lens.
3. Calculating the focal length [tex]\( f \)[/tex] of the lens:
- Using the lens formula:
[tex]\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\][/tex]
- Substituting [tex]\( v = -100 \)[/tex] cm and [tex]\( u = 50 \)[/tex] cm:
[tex]\[
\frac{1}{f} = \frac{1}{-100} + \frac{1}{50}
\][/tex]
[tex]\[
\frac{1}{f} = -\frac{1}{100} + \frac{2}{100}
\][/tex]
[tex]\[
\frac{1}{f} = \frac{1}{100}
\][/tex]
[tex]\[
f = 100 \text{ cm} = 1 \text{ m}
\][/tex]
4. Calculating the power [tex]\( P \)[/tex] of the lens:
- The power [tex]\( P \)[/tex] of a lens is given by:
[tex]\[
P = \frac{1}{f(\text{in meters})}
\][/tex]
- Substituting [tex]\( f = 1 \)[/tex] m:
[tex]\[
P = \frac{1}{1} = 1 \text{ D}
\][/tex]
Summary:
- The nature of the lenses:
1. +2D: Converging (Convex) lens.
2. -4D: Diverging (Concave) lens.
- The object should be placed 50 cm in front of the lens to get the required image size.
- The power of the lens is 1 diopter (1D).
