Tag: reflection of light by curved surfaces

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Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

A $10\mathrm { mm }$ long awl pin is placed vertically In front of a concave mirror. A $5\ mm$ long image of the awl pin is formed at $30\mathrm { cm }$ in front of the mirror, The focal length of this mirror is 

  1. $- 30 \mathrm { cm }$
  2. $- 20 \mathrm { cm }$
  3. $- 40 \mathrm { cm }$
  4. $- 60 \mathrm { cm }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnification m = h_i/h_o = -v/u. m = -5/10 = -0.5. -0.5 = -(-30)/u => u = -60. Mirror formula: 1/f = 1/v + 1/u = 1/-30 + 1/-60 = -3/60 = -1/20. f = -20 cm.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

A point object is placed on principal axis of concave mirror of radius of curvature 10 cm at a distance 21 cm from pole of the mirror.  A glass slab of thickness 3 cm and refractive index 1.5 is placed between object and mirror $.$ Find the imaged position of the image formed. 

  1. 4

  2. 3

  3. 16.5

  4. 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

A converging bundle of light rays in the shape of cone with a vertex angle of 45 falls on a circular diaphragm of 20 cm diameter. A lens with power 5 D is fixed in the diaphragm. Diameter of face of lens is equal to that of diaphragm. If the vertex angle of new cone is 

  1. $
    \cfrac { 3 d } { 4 }
    $
  2. $
    \cfrac { 5 d } { 4 }
    $
  3. $
    2 d
    $
  4. $
    \cfrac { 3 } { 2 } d
    $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

All the following statements are correct except

  1. The magnification produced by a convex mirror is always less than one.

  2. A virtual, crect , same-size image can be obtained by using a plane mirror.

  3. A virtual, crect , magnified image can be formed by using a convex mirror.

  4. A virtual, inverted , same-sized image can be formed by using a convex mirror.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A convex mirror always forms a virtual, erect, and diminished image for any real object position. Therefore, the magnification is always less than 1. Statement A is correct. Statement D is incorrect because a convex mirror cannot form an inverted image for a real object.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

The suns diameter is $1.4\times { 10 }^{ 9 }m$ and its distance from the earth is ${ 10 }^{ 11 }m$. The diameter of its image, formed by a convex mirror of focal length 2m will

  1. 0.7 cm

  2. 1.4 cm

  3. 2.8 cm

  4. 10 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a mirror, the magnification m = -v/u = f/(f-u). Given the sun is very far away (u is large), the image is formed at the focus f. The size of the image is h_i = h_o * (f/u). Plugging in values: (1.4 * 10^9) * (2 / 10^11) = 2.8 * 10^-2 m = 2.8 cm.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

The distance of real object when a concave mirror produces a real image of magnification $'m'$ is ($f$ is focal length)

  1. $\left(\frac{m - 1}{m}\right) f$
  2. $\left(\frac{m + 1}{m}\right) f$
  3. $(m-1)f$
  4. $(m+1)f$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a concave mirror, magnification m = f / (f - u). Rearranging for u: m(f - u) = f, mf - mu = f, mu = mf - f, u = f(m - 1) / m. However, for a real image, m is negative. Using m = -|m|, the distance u = f(1 + |m|) / |m|.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

Converging rays are incident on a convex spherical mirror so that their extensions intersect  $30 cm$  behind the mirror on the optical axis. The reflected rays form a diverging beam, so that their extensions intersect the optical axis  $1.2 m$  from the mirror. The focal length of the mirror is

  1. $40{ cm }$
  2. $60{ cm }$
  3. $30{ cm }$
  4. $24{ cm }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the mirror formula 1/v + 1/u = 1/f. For converging rays, u = +30 cm (virtual object). The reflected rays form a diverging beam with image at v = -120 cm. 1/(-120) + 1/30 = 1/f. 1/f = (-1 + 4) / 120 = 3/120 = 1/40. So f = 40 cm.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

Magnification produced by a rear view mirror fitted in vehicles

  1. is less than one

  2. is more than one

  3. is equal to one

  4. can be more than or less than one depending upon the position of the object in front of it.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rear view mirrors in vehicles are convex mirrors. They are designed to provide a wider field of view by forming diminished images, so the magnification is always less than one.

Multiple choice physics reflection of light in spherical mirrors focus and focal length spherical mirror formula and magnification reflection of light by curved surfaces

A concave mirror of radius of curvature 40 cm forms an image of an object placed on the principal axis at a distance 45 cm in front of it. Now if the system (including object) is completely immersed in water $(\mu=1.33)$, then:

  1. the image will shift towards the mirror.

  2. the magnification will reduce.

  3. the image will shift away from the mirror and magnification will increase.

  4. the position of the image and magnification will not change.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Mirror operates on the principle of Laws of reflection. 

Therefore, focal length does not depend upon the medium and object distance is also unchanged, so their will be no change in the image distance, correspondingly magnification will also remain unchanged.