An object of size 7.0cm is placed at a distance of 27cm in front of concave mirror of focal length 18cm. At what distance from the mirror should a screen be placed so that a sharply focused image can be obtained? Find the size and the nature of the image.
Download our app for free and get started
Given; Focal length of concave mirror (f)= -18cm Object distance (u)= -27cm Object height= 7cm Let image distance be vcm We know by mirror formula that$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}+\frac{1}{(-27)}=\frac{1}{(-18)}$
$\frac{1}{\text{v}}=\frac{1}{(-18)}+\frac{1}{27}$
$\frac{1}{\text{v}}=\frac{(-3+2)}{54}$
$\frac{1}{\text{v}}=\frac{-1}{54}$
$\text{v}=-54\text{cm}$
Thus, placing a screen at a distance of 54cm from the concave mirror on the same side as object will give us a sharp and focused image. The magnification will be$\text{m}=-\frac{\text{v}}{\text{u}}$
$\text{m}=-\frac{(-54)}{(-27)}$
$\text{m}=-2$
|m|>1 ⇒ enlarged image and the negative sign implies that the image is inverted. m can also be written as$\text{m}=\frac{\text{Height of image}}{\text{Height of object}}$
$-2=\frac{\text{Height of image}}{7}$
Height of image= -14cm Thus the image has a height of 14cm and is inverted and present at a distance of 54cm from the mirror.
$\frac{1}{\text{v}}+\frac{1}{(-27)}=\frac{1}{(-18)}$
$\frac{1}{\text{v}}=\frac{1}{(-18)}+\frac{1}{27}$
$\frac{1}{\text{v}}=\frac{(-3+2)}{54}$
$\frac{1}{\text{v}}=\frac{-1}{54}$
$\text{v}=-54\text{cm}$
Thus, placing a screen at a distance of 54cm from the concave mirror on the same side as object will give us a sharp and focused image. The magnification will be$\text{m}=-\frac{\text{v}}{\text{u}}$
$\text{m}=-\frac{(-54)}{(-27)}$
$\text{m}=-2$
|m|>1 ⇒ enlarged image and the negative sign implies that the image is inverted. m can also be written as$\text{m}=\frac{\text{Height of image}}{\text{Height of object}}$
$-2=\frac{\text{Height of image}}{7}$
Height of image= -14cm Thus the image has a height of 14cm and is inverted and present at a distance of 54cm from the mirror.
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 SolutionAn object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
- 2View SolutionWith the help of a ray diagram, state the meaning of refraction of light. State Snell’s law of refraction of light and also express it mathematically.
- 3View SolutionDraw a ray diagram showing the path of rays of light when it enters with oblique incidence:
- From air into water;
- From water into air.
- 4View SolutionBetween which two points of concave mirror should an object be placed to obtain a magnification of:
- −3
- +25
- −0.4
- 5View 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.
- 6View SolutionWhy does a ray falling normally on a plane mirror, retrace its path?
- 7The refractive indices of water and glass are $\frac{3}{4}\text{ and }\frac{3}{2}$ respectively. Write the relation and find the value of refractive index of water with respect to glass and glass with respect to water.View Solution
- 8View SolutionThe image of an object formed by a lens is of magnification - 1. If the distance between the object and its image is 60 cm, what is the focal length of the lens? If the object is moved 20 cm towards the lens, where would the image be formed? State reason and also draw a ray diagram in support of your answer.
- 9View SolutionDescribe the New Cartesian Sign Convention used in optics. Draw a labelled diagram to illustrate this sign convention.
- 10View SolutionA concave mirror of focal length 1.5m forms an image of an object placed at a distance of 40cm. Find the position and nature of the image.