A 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.
Download our app for free and get started
According to the question; Object distance = -30cm; Image distance = vcm; Focal length = 20cm By lens formula;$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{-30}=\frac{1}{20}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{20}-\frac{1}{30}$
$\Rightarrow\frac{1}{\text{v}}=\frac{30-20}{600}=\frac{10}{600}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{60}$
$\Rightarrow\text{v}=60\text{cm.}$
Therefore image is formed at 60cm in right of lens. Now; Height of object Height of object $h_1=5 cm$;; Magnification $=\frac{\text{h}_2}{\text{h}_1}=\frac{\text{v}}{\text{u}}$ Putting values of v and u Magnification $=\frac{\text{h}_2}{5}=\frac{60}{-30}$$\Rightarrow\frac{\text{h}_2}{5}=-2$
$\Rightarrow\text{h}_2=-2\times5=10$
Height of image is 10cm. Negative sign means image is real and inverted.
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{-30}=\frac{1}{20}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{20}-\frac{1}{30}$
$\Rightarrow\frac{1}{\text{v}}=\frac{30-20}{600}=\frac{10}{600}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{60}$
$\Rightarrow\text{v}=60\text{cm.}$
Therefore image is formed at 60cm in right of lens. Now; Height of object Height of object $h_1=5 cm$;; Magnification $=\frac{\text{h}_2}{\text{h}_1}=\frac{\text{v}}{\text{u}}$ Putting values of v and u Magnification $=\frac{\text{h}_2}{5}=\frac{60}{-30}$$\Rightarrow\frac{\text{h}_2}{5}=-2$
$\Rightarrow\text{h}_2=-2\times5=10$
Height of image is 10cm. Negative sign means image is real and inverted.
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 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?
- 2View 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?
- 3View SolutionA lens forms a real image 3cm high of an object 1cm high. If the separation of object and image is 15cm, find the focal length of the lens.
- 4View SolutionAn object of 5.0cm size is placed at a distance of 20.0cm from a converging mirror of focal length 15.0cm. At what distance from the mirror should a screen be placed to get the sharp image? Also calculate the size of the image.
- 5View SolutionThe image of a candle flame placed at a distance of 45cm from a spherical lens is formed on a screen placed at a distance of 90cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2cm, find the height of its image.
- 6View SolutionAn object is placed at a distance of 100cm from a converging lens of focal length 40cm.
- What is the nature of image?
- What is the position of image?
- 7View SolutionAt what distance from a concave mirror of focal length 10cm should an object 2cm long be placed in order to get an erect image 6cm tall?
- 8View SolutionA small object is placed 150mm away from a diverging lens of focal length 100mm:
- Copy the figure below and draw rays to show how an image is formed by the lens.
- Calculate the distance of the image from the lens by using the lens formula.
- 9View SolutionA convex lens of focal length 6cm is held 4cm from a newspaper which has print 0.5cm high. By calculation, determine the size and nature of the image produced.
- 10View SolutionTo construct a ray diagram we use two rays of light which are so chosen that it is easy to determine their directions after reflection from the mirror. Choose these two rays and state the path of these rays after reflection from a concave mirror. Use these two rays to find the nature and position of the image of an object placed at a distance of 15cm from a concave mirror of focal length 10cm.