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Questions Related to physics

Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

An object of length 2.0 cm is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. Find the size of the image if the object is at a distance of 8.0 cm from the lens.

  1. $6 cm$
  2. $4 cm$
  3. $5 cm$
  4. $1 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using 1/v - 1/u = 1/f with u = -8 cm and f = 12 cm, 1/v = 1/12 - 1/8 = (2-3)/24 = -1/24. So v = -24 cm. Magnification m = v/u = -24 / -8 = 3. Image size = m * object size = 3 * 2.0 cm = 6 cm.

Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

A point object $O$ is placed on the principle axis of a convex lens of focal length $20\ cm$ at a distance of $40\ cm$ to the left of it. The diameter of the lens is $10\ cm$. If the eye is placed $60\ cm$ to the right of the lens at a distance $h$ below the principle axis, then the maximum value of $h$ to see the image will be

  1. $2.5\ cm$
  2. $5\ cm$
  3. $0\ cm$
  4. $10\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The object is at 2f (40 cm), so the image is at 2f (40 cm) on the other side. The lens diameter is 10 cm (radius 5 cm). The rays from the object pass through the lens and converge at the image point. The cone of light has a radius that scales with distance. At 60 cm from the lens (20 cm past the image), the cone radius is determined by similar triangles.

Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

For two position of lens, the images are obtained on a fixed screen. If the size of the object is 2 cm and size of diminished image is 0.5 cm, the size of the other image will be

  1. 1 cm

  2. 4 cm

  3. 8 cm

  4. 16 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$height^{2} _{object}=height _{image1}\times height _{image2}$


$2 \times 2=0.5 \times h$

$h=8$

option $C$ is correct

Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

A lens is placed between the source of light and a wall. It forms images of area $ { A } _{ 1 }$ and  ${ A } _{ 2 }$ on the wall for its two different positions. The area of the source of light is :

  1. $ \sqrt { { A } _{ 1 }{ A } _{ 2 } } $
  2. $ \dfrac { { A } _{ 1 }+{ A } _{ 2 } }{ 2 } $
  3. $ { \left( \dfrac { \sqrt { { A } _{ 1 } } +\sqrt { { A } _{ 2 } } }{ 2 } \right) }^{ 2 } $
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$height^{2} _{object}=height _{image1} \times height _{image2}$


$r^{2} _{source}=r _{image1} \times r _{image2}$

$\pi r^{2} _{source}=\pi r _{image1} \times r _{image2}$

$A _{source}=\sqrt{\pi^{2} r^{2} _{image1}r^{2} _{image2}}$

$A _{source}=\sqrt{A _{1}A _{2}}$

option $A$ is correct 

Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

A student focused the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under position of candle $=12.0\ cm$position of convex lens $=50.0\ cm$position of the screen $=88.0\ cm$. Where will the image be formed, if he shifts the candle towards the lens at a position of $31.0\ cm$?

  1. 19cm

  2. 48cm

  3. Infinity

  4. at center of curvature
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial positions: candle 12, lens 50, screen 88. Object distance u = 50 - 12 = 38 cm. Image distance v = 88 - 50 = 38 cm. Since u = v, 2f = 38, so f = 19 cm. If the candle is moved to 31 cm, the new object distance u = 50 - 31 = 19 cm. Since u = f, the image is formed at infinity.