A convex lens forms a real and inverted image of an object. Where is the needle placed in front of the convex lens in relation to the size of the object? Also, find the power of th... A convex lens forms a real and inverted image of an object. Where is the needle placed in front of the convex lens in relation to the size of the object? Also, find the power of the lens. Find the power of a concave lens with a focal length of 2m. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance from the object? What is the nature of the image? Draw a ray diagram to show the image formation.
Understand the Problem
The question involves optics and asks about the formation of images by a convex lens and a concave mirror. It requires calculations for the lens's power and descriptions of image characteristics based on object placement.
Answer
- Needle is placed 25 cm in front of the convex lens; power of convex lens is $4 \text{ diopters}$; power of concave lens is $-0.5 \text{ diopters}$; for concave mirror, object distance must be less than $15 \text{ cm}$ for a virtual image.
Answer for screen readers
- Needle is placed 25 cm in front of the convex lens.
- Power of convex lens = 4 diopters.
- Power of concave lens = -0.5 diopters.
- For the concave mirror, place the object less than 15 cm from the mirror. The image is virtual and erect.
Steps to Solve
- Finding Object Distance for the Convex Lens
To find where the needle is placed in front of the convex lens, we use the lens formula:
$$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $$
Given values:
- Image distance, $v = 50 \text{ cm}$
- Focal length, $f$ is calculated using the fact that the image distance is twice the focal length ($v = 2f$), hence
$$ 2f = 50 \implies f = 25 \text{ cm} $$
Now, substitute the values into the lens formula:
$$ \frac{1}{25} = \frac{1}{50} - \frac{1}{u} $$
To find $u$ (object distance), we rearrange and solve:
$$ \frac{1}{u} = \frac{1}{50} - \frac{1}{25} $$
- Calculating Object Distance
Calculating right-hand side gives:
$$ \frac{1}{u} = \frac{2 - 1}{50} = \frac{1}{50} $$
Thus,
$$ u = 50 \text{ cm} $$
The needle is placed 25 cm in front of the convex lens.
- Finding the Power of the Convex Lens
The power of a lens can be calculated using the formula:
$$ P = \frac{1}{f \text{ (in meters)}} $$
Focal length $f = 0.25 \text{ m}$, thus:
$$ P = \frac{1}{0.25} = 4 \text{ diopters} $$
- Finding the Power of the Concave Lens
For a concave lens with focal length $f = -2 \text{ m}$ (negative since it's concave):
$$ P = \frac{1}{f} = \frac{1}{-2} = -0.5 \text{ diopter} $$
- Finding the Range of Distance for the Concave Mirror
For an erect image with a concave mirror of focal length $f = -15 \text{ cm}$, use the mirror formula:
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
To have an erect image, the object must be placed between the focal point and the mirror. Thus, the object distance is less than 15 cm.
- Nature of the Image
The image formed by a concave mirror when the object is placed between the focal point and the mirror is virtual and upright.
- Sketching the Ray Diagram
Draw a concave mirror, the focus (F), and the center of curvature (C). Plot the object between F and the mirror, showing light rays reflecting and forming a virtual image.
- Needle is placed 25 cm in front of the convex lens.
- Power of convex lens = 4 diopters.
- Power of concave lens = -0.5 diopters.
- For the concave mirror, place the object less than 15 cm from the mirror. The image is virtual and erect.
More Information
The convex lens produces real and inverted images primarily located beyond twice its focal length. In contrast, the concave mirror creates a virtual image when the object is within its focal length.
Tips
- Confusing the sign conventions (positive for convex and negative for concave lenses/mirrors).
- Mixing up real and virtual images, mainly regarding object placement relative to the focal point.
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