Tag: problems on mirror and magnification formula

Questions Related to problems on mirror and magnification formula

If an object is placed at a distance of 20cm from the pole of a concave mirror, the magnification of its real image is 3. If the object is moved away from the mirror by 10cm, then the magnification is -1.

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. True

  2. False

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Correct Option: A
Explanation:

$M= \frac{f}{f-d _0}$ and real image has M negative

$-3= \frac{f}{f-20}$

$-3f+60=f$

$f=15 cm$

$M= \frac{15}{15-30}$

$M= -1$

A convex lens is given, for which the minimum distance between an object and its rel image is $40cm$. An object is placed at a distance of $15cm$ from this lens. The liner magnification of adjustment will be 

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $\dfrac{5}{3}$

  2. $-2$

  3. $2$

  4. $\dfrac{1}{2}$

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Correct Option: A
Explanation:

Given object distance $u=15$ cm

Distance between object and real image produced $=40 $cm
Thus image distance $v=40-15=25$ cm
Also we know linear magnification,
$m=\dfrac{-v}{u}=\dfrac{-25}{-15}=\dfrac{5}{3}$ 

An object is placed in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm, the position and nature of the image formed is :

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $+30 cm, virtual \ and \ erect$

  2. $+30 cm, real \ and \ inverted$

  3. $-30 cm, virtual \ and \ erect$

  4. $-30 cm, real \ and \ inverted$

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Correct Option: D

An object of length $6\ cm$ is placed on the principle axis of a concave mirror of focal length $f$ at a distance of $4\ f$. The length of the image will be

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $2\ cm$

  2. $12\ cm$

  3. $4\ cm$

  4. $1.2\ cm$

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Correct Option: A
Explanation:

Given that,

The object distance $u=-6\,cm$

Now, magnification is

  $ m=\dfrac{I}{O} $

 $ m=\dfrac{f}{f-u} $

 $ \dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)} $

 $ I=-2\,cm $

Hence, the length of image is -$2\ cm$

An astronomical telescope has focal lengths $100$ & $10$cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision,magnification power of telescope will be,

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. -15

  2. -14

  3. -17

  4. -19

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Correct Option: B
Explanation:

Given focal length of eye piece${f} _{e}=10cm\$

focal length of objective${f} _{o}=100 cm\$
Also we know least distance $D=25 cm\$ 
Magnifying power $M=\dfrac{-{f} _{0}}{{f} _{e}}(1+\dfrac{{f} _{e}}{D})\$
$M=-\dfrac{100}{10}(1+\dfrac{10}{25})\$
$M=-14$

In the displacement method, a convex lens is placed in between an object and a screen. If one of the magnification is $3$ and the displacement of the lens between the two positions is $24$cm, then the focal length of the lens is:

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $10$ cm

  2. $9$ cm

  3. $6$ cm

  4. $16/3$ cm

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Correct Option: B
Explanation:

Given magnification $M=\dfrac{v}{u}=3$


Thus $v=3u$, where v and u are the image and object distance respectively.

Also Distance between lenses$=v-u=24$
Thus $u=12 cm$, than $v=36 cm$

From lens formula we have,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{f}=\dfrac{1}{36}+\dfrac{1}{12}$

$\dfrac{1}{f}=\dfrac{4}{36}$

$f=\dfrac{36}{4}$

$f=9 cm$

A concave mirror of focal length $20\ cm$ produces an image twice the height of the object. If the image is real, then the distance of the object from the mirror is:

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $20\ cm$

  2. $60\ cm$

  3. $10\ cm$

  4. $30\ cm$

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Correct Option: D

In a concave mirror an object is placed at a distance x from the focus, and the image is formed at a distance y from the focus. The focal length of the mirror is

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $xy$

  2. $\sqrt{xy} $

  3. $\dfrac{x+y}{2} $

  4. $\sqrt{\dfrac{x}{y} }$

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Correct Option: B

Sun subtends an angle of $0.5^{o}$ at the pole of a concave mirror of radius of curvature 15 m. The diameter of the image of the sun formed by the mirror is

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $8.55 cm$

  2. $7.55 cm$

  3. $6.55 cm$

  4. $5.55 cm$

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Correct Option: A

A light ray travelling parallel to the principle axis of a concave mirror strikes the minor at angle of incidence $\theta$. If the radius of curvature of the mirror is $R$, then after reflection, the ray meets the principle axis at distance $d$ from the centre of curvature, then $d$ is 

physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

  1. $\dfrac {R}{2}$

  2. $R\left(1-\dfrac {1}{2\cos \theta}\right)$

  3. $\dfrac {R}{2\cos \theta}$

  4. $\dfrac {R}{2}(1+\cos \theta)$

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Correct Option: A