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# a lens produces an image of magnification 2 which is virtual and erect. if the focal length of the lens is 100 mm. the position of the object will be

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get a lens produces an image of magnification 2 which is virtual and erect. if the focal length of the lens is 100 mm. the position of the object will be from screen.

## The magnifications produced by a convex lens for two different positions of an object are m1 and m2 respectively ( m1 > m2 ) . If d is the distance of separation between the two positions of the object then the focal length of the lens is

Click hereπto get an answer to your question βοΈ The magnifications produced by a convex lens for two different positions of an object are m1 and m2 respectively ( m1 > m2 ) . If d is the distance of separation between the two positions of the object then the focal length of the lens is

The magnifications produced by a convex lens for two different positions of an object are m

Question 1 β and m 2 β respectively (m 1 β > m 2 β

). If d is the distance of separation between the two positions of the object then the focal length of the lens is

A

m 1 β m 2 β β

B

m 1 β βm 2 β β d β

C

m 1 β βm 2 β dm 1 β m 2 β β

D

m 1 β βm 2 β d β Medium Open in App Solution Verified by Toppr

Correct option is D)

Given: The magnifications produced by a convex lens for two different positions of an object are m

1 β and m 2 β respectively (m 1 β >m 2 β

). If d is the distance of separation between the two positions of the object

To find the focal length of the lens

Solution:

Separation between object and image for first position,

D=v+u

where u,v are the object distance and image distance, respectively.

So, magnification m 1 β = u v β βΉv=um 1 β ....(i)

When the lens is at second position,

d=v-u So, m 2 β = v u β βΉu=vm 2 β ......(ii) So, m 1 β m 2 β =1

From lens equation and using eqn (i), we get

f 1 β = v 1 β + u 1 β βΉ f 1 β = um 1 β 1 β + u 1 β βΉ f 1 β = um 1 β 1+m 1 β β βΉu= m 1 β f(m 1 β +1) β .........(iii)

From lens equation and using eqn (ii), we get

f 1 β = v 1 β + u 1 β βΉ f 1 β = vm 2 β 1 β + v 1 β βΉ f 1 β = vm 2 β 1+m 2 β β βΉv= m 2 β f(m 2 β +1) β .........(iv)

substituting the values of eqn(iii) and(iv) in the following equation, we get

d=vβu βΉd= m 2 β f(m 2 β +1) β β m 1 β f(m 1 β +1) β βΉd=f( m 1 β m 2 β m 1 β (m 2 β +1)βm 2 β (m 1 β +1) β ) βΉf= m 1 β m 2 β +m 1 β βm 1 β m 2 β βm 2 β dm 1 β m 2 β β βΉf= m 1 β βm 2 β dm 1 β m 2 β β but m 1 β m 2 β =1

Therefore the focal length becomes,

f= m 1 β βm 2 β d β

20 16

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## [Solved] If the magnification produced by a lens is +2, then the imag

CONCEPT:β It is also a ratio between the distance of the image to the distance of the object from the lens. $$\Rightarrow m=\frac{h_i}{h_o}=\frac{v}{u}$$

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## If the magnification produced by a lens is +2, then the image is:

This question was previously asked in

CDS 01/2022: GK Previous Paper (Held On 10 April 2022)

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erect, virtual and smaller than the object.

inverted, real and smaller than the object.

erect, virtual and larger than the object.

inverted, real and larger than the object.

Option 3 : erect, virtual and larger than the object.

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## Detailed Solution

CONCEPT:β

It is also a ratio between the distance of the image to the distance of the object from the lens.

βm=hiho=vu

If magnification is more than 1, the image is magnified and if it is less than one image is diminished.

If magnification is positive, the image is erect, virtual and if it is negative, the image is inverted, real.

EXPLANATION:

βMagnification, m = +2 represents the case of an Enlarged image and hence it must be a Convex Lens.

The image formed by a Convex lens is Erect and Virtual.

Hence the image is erect, virtual, and larger than the object.

CONVEX LENS CONCAVE LENS

Coverges all the light rays at a single point.  Diverges all the light rays from a single point.

Thicker from the middle and thinner from the sides. Thicker at the ends and thinner from the middle.

Magnification may be Positive or Negative Magnification is always Positive.

An image formed is erect and virtual. Enlarged if the object is placed between lens and 2F otherwise Diminished.  An image formed is inverted, virtual, and diminished

Important Note: A Positive magnification shows the image is virtual and erect (upright) and a negative one shows the image is real and inverted.

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## More Refraction and Reflection Questions

Q1. A convex lens has a focal length of 15 cm. At what distance should an object be placed in front of the lens to get a real image of the same size of the object?Q2. The twinkling of a star is due to:Q3. If the magnification produced by a lens is +2, then the image is:Q4. What is the magnification produced by a concave lens of focal length 10 cm, when an image is formed at a distance of 5 cm from the lens?Q5. The twinkling of a star is due to the atmosphericQ6. _______ are commonly used as rear view mirrors in vehicles.Q7. An object, 3.0 cm in height, is placed at a distance of 3.0 m in front of a convex mirror of focal length 1.5 m, on its principal axis. Following New Cartesian Sign Convention, the image is formed at v =___________ and its height hI =_________ .Q8. Two lenses Lβ and Lβ of the same power, +5.0 D, separated by a distance of 90 cm are arranged parallel to each other in vertical planes. An object is placed left to Lβ at a distance of 30 cm, on the common axis of the lenses. The distance between the original object and the final image is:Q9. Which of the following statements about the image formed by a concave lens is always correct?Q10. An object is placed on the principal axis of a convex lens at a point between F1 and 2 F1 (F1 is the principal focus to the left of the lens). The image formed is:

## More Optics Questions

Q1. A convex lens has a focal length of 15 cm. At what distance should an object be placed in front of the lens to get a real image of the same size of the object?Q2. The twinkling of a star is due to:Q3. If the magnification produced by a lens is +2, then the image is:Q4. What is the magnification produced by a concave lens of focal length 10 cm, when an image is formed at a distance of 5 cm from the lens?

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## A concaye mirror of focal length 18 cm produces 3 times magnified errect image of an object. Find the position of the object.

For concave mirror, focal length, f = -18 cm Magnification, m=+3 m=-(v)/(u)=(h(i))/(h(o))=3 impliesv=-3u Applying mirror formula, (1)/(v)+(1)/(u)=(1)/(f) (1)/(-3u)+(1)/(u)=(1)/(-18) u=-12cm Hence, position of object is -12 cm.

Home > English > Class 12 > Physics > Chapter >

Ahsec Examination Paper 2020 (Fully Solved) (Physics Xii)

>

A concaye mirror of focal leng...

A concaye mirror of focal length 18 cm produces 3 times magnified errect image of an object. Find the position of the object.

Text Solution Open Answer in App Solution

For concave mirror, focal length, f = -18 cm Magnification, m=+3

m=β v u = h i h o =3 m=-vu=hiho=3 βv=β3u βv=-3u

Applying mirror formula,

1 v + 1 u = 1 f 1v+1u=1f 1 β3u + 1 u = 1 β18 1-3u+1u=1-18 u=β12cm u=-12cm

Hence, position of object is -12 cm.

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