Size of image of an object by a mirror having a focal length of 20cm is observed to be reduced to $\frac{1}{3}\text{rd}$ of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?
Download our app for free and get started
Since image size is $\frac{1}{3}$ of object size, so image distance is $\frac{1}{3}$ of object distance because $\frac{\text{h}'}{\text{h}}=\frac{\text{v}}{\text{u}}$ Using the mirror formula, we can calculate object distance and image distance,$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
or, $\frac{3}{\text{u}}-\frac{1}{\text{u}}=-\frac{1}{20}$ or, $\frac{3-1}{\text{u}}=-\frac{1}{20}$ or, $\text{u}=-40\text{cm}$ Using this, we can find image distance:$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
or, $\frac{1}{\text{v}}-\frac{1}{40}=-\frac{1}{20}$ or, $\frac{1}{\text{v}}=-\frac{1}{20}+\frac{1}{40}$ or, $\frac{-2+1}{40}=-\frac{1}{40}$But above value of image distance does not match with our initial assumption. This means that the mirror is not a concave mirror but a convex mirror. Let us calculate with the assumption that it is a convex mirror.
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
or, $\frac{1}{\text{v}}-\frac{1}{40}=-\frac{1}{20}$ or, $\frac{1}{\text{v}}=-\frac{1}{20}+\frac{1}{40}=\frac{3}{40}$ or, $\text{v}=\frac{40}{3}\text{cm}$Nature of mirror: convex mirror.
Nature of image: Smaller than object, erect and virtual.
or, $\frac{3}{\text{u}}-\frac{1}{\text{u}}=-\frac{1}{20}$ or, $\frac{3-1}{\text{u}}=-\frac{1}{20}$ or, $\text{u}=-40\text{cm}$ Using this, we can find image distance:$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
or, $\frac{1}{\text{v}}-\frac{1}{40}=-\frac{1}{20}$ or, $\frac{1}{\text{v}}=-\frac{1}{20}+\frac{1}{40}$ or, $\frac{-2+1}{40}=-\frac{1}{40}$But above value of image distance does not match with our initial assumption. This means that the mirror is not a concave mirror but a convex mirror. Let us calculate with the assumption that it is a convex mirror.
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
or, $\frac{1}{\text{v}}-\frac{1}{40}=-\frac{1}{20}$ or, $\frac{1}{\text{v}}=-\frac{1}{20}+\frac{1}{40}=\frac{3}{40}$ or, $\text{v}=\frac{40}{3}\text{cm}$Nature of mirror: convex mirror.
Nature of image: Smaller than object, erect and virtual.
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1View SolutionHow is the refractive index of a medium related to the speed of light? Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media?
- 2View SolutionDescribe an activity to show that the colours of white light splinted by a glass prism can be recombined to get white light by another identical glass prism. AIso draw ray diagram to show the recombination of the spectrum of white light.
- 3View SolutionThe image of an object place at 60cm in front of a lens is obtained on a screen at a distance of 120cm from it. Find the focal length of the lens. What would be the lens? What would be the height of the image? If the object is 5cm high?
- 4View SolutionA dentist uses a mirror in front of a decayed tooth at a distance of 4cm from the tooth to get four times magnified image in the mirror. Use mirror formula to find the focal length and nature of the mirror used.
- 5View SolutionMake labelled ray diagrams to illustrate the formation of:
- A real image by a converging mirror.
- A virtual image by a converging mirror.
- 6View SolutionWith the help of a diagram, explain why the image of an object viewed through a concave lens appears smaller and closer than the object.
- 7View SolutionA 5.0cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20cm. The distance of the object from the lens is 30cm By calculation determine.
- The position and
- The size of the image formed.
- 8View SolutionCalculate the image distance for an object of height 12mm at a distance of 0.20m from a concave lens of focal length 0.30 m, and state the nature and size of the image.
- 9View SolutionAn erect image 2.0cm high is formed 12cm from a lens, the object being 0.5cm high. Find the focal length of the lens.
- 10View SolutionA ray of light enters a diamond from air. If the refractive index of diamond is 2.42, what % (percentage) is speed of light in diamond with that of speed in air?