An object 50cm tall is placed on the principal axis of a convex lens. Its 20cm tall image is formed on the screen placed at a distance of 10cm from the lens. Calculate the focal length of the lens.
Download our app for free and get started
Height of object (h) = 50cm Height of image (h') = -20cm (real and inverted) Distance of image from the lens (v) = 10cm Distance of object from the lens (u) = ? Focal length of the lens (f) = ? We know, magnification (m) of the lens is given by,$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}'}{\text{h}}$
Thus, substituting the values of v, h and h', we get,$\frac{10}{\text{u}}=\frac{-20}{50}$
${\text{u}}=\frac{-5}{2}\times10$
$\therefore\ \text{u}=-25\text{cm}$
Using the lens formula,$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{10}-\frac{1}{-25}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{10}+\frac{1}{25}=\frac{1}{\text{f}}$
$\Rightarrow\frac{5+2}{50}=\frac{1}{\text{f}}$
$\Rightarrow\frac{7}{50}=\frac{1}{\text{f}}$
$\Rightarrow\text{f}=\frac{50}{7}$
$\Rightarrow\text{f}=7.14\text{cm}$
Thus, substituting the values of v, h and h', we get,$\frac{10}{\text{u}}=\frac{-20}{50}$
${\text{u}}=\frac{-5}{2}\times10$
$\therefore\ \text{u}=-25\text{cm}$
Using the lens formula,$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{10}-\frac{1}{-25}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{10}+\frac{1}{25}=\frac{1}{\text{f}}$
$\Rightarrow\frac{5+2}{50}=\frac{1}{\text{f}}$
$\Rightarrow\frac{7}{50}=\frac{1}{\text{f}}$
$\Rightarrow\text{f}=\frac{50}{7}$
$\Rightarrow\text{f}=7.14\text{cm}$
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 SolutionDefine power of a lens. What is its unit? One student uses a lens of focal length 50cm and another of -50cm. What is the nature of the lens and its power used by each of them?
- 2View SolutionA concave mirror produces three times enlarged virtual image of an object placed at 10cm in front of it. Calculate the radius of curvature of the mirror.
- 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 SolutionTwo lenses A and B have focal lengths of +20cm and, -10cm, respectively.
- What is the nature of lens A and lens B?
- What is the power of lens A and lens B?
- What is the power of combination if lenses A and B are held close together?
- 5View SolutionIf the image formed by a lens for all positions of an object placed in front of it is always erect and diminished, what is the nature of this lens? Draw a ray diagram to justify your answer. If the numerical value of the power of this lens is 10 D, what is its focal length in the Cartesian system?
- 6View SolutionHow far should an object be placed from a convex lens of focal length 20cm to obtain its image at a distance of 30cm from the lens? What will be the height of the image, if the objects is 6cm tall?
- 7View SolutionDetermine how far an object must be placed in front of a converging lens of focal length 10cm in order to produce an erect (upright) image of linear magnification 4.
- 8View SolutionThe optical prescription for a pair of spectacles is:
Right eye: 3.50D, Left eye: 4.00D- Are these lenses thinner at the middle or at the edges?
- Which lens has a greater focal length?
- Which is the weaker eye?
- 9View SolutionAn object of height 4cm is placed at a distance of 15cm in front of a concave lens of power, -10 dioptres. Find the size of the image.
- 10View SolutionThe two rays chosen for the construction of ray diagram is:
Ray 1: When the incident ray is parallel to the principal axis, the reflected ray will pass through the focus of concave mirror or it appears to pass through the focus of convex mirror.
Ray 2: When the incident ray passes through or appears to pass through the centre of curvature, the light, after reflection from the spherical mirror, reflects back along the same path.
The image formed is real, inverted, magnified and is formed beyond the centre of curvature.
