Physics

Optics and Lenses

251 Questions

Enhance your physics knowledge by practicing questions on optics and the functioning of lenses. The set covers calculating combined focal lengths, lens power, and correcting vision defects like myopia and hypermetropia. These physics fundamentals are crucial for medical entrance and state board exams.

Combined focal lengthCorrecting vision defectsLens power calculationConvex and concave lensesTelescopesHuygens principle

Optics and Lenses Questions

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

An object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is obtained on the screen. The linear magnification of the image is 25. The lens is now moved 30 cm towards the screen and a sharp image is again formed on the screen. Find the focal length of the lens.

  1. $1.2 cm$
  2. $14.3 cm$
  3. $14.6 cm$
  4. $14.9 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the displacement method formulas: m1 = 25, m2 = 1/25 (since the lens is moved). The distance between positions is d = 30 cm. The formula for focal length is f = (D^2 - d^2) / 4D. Alternatively, using magnification m = (D-d)/2f, one can solve for f.

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

A convex lens forms an image of an object on a screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90 cm.

  1. The distance between the two positions of the lens is 30 cm.

  2. The distance of the object from the lens in its first position is 36 cm.

  3. The height of the object is 6 cm.

  4. The focal length of the lens is 21.6 cm.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


$h _{object}=\sqrt{36}=6$

magnification of image is $\dfrac{v}{u}=\dfrac{9}{6}$

                                            $v= \dfrac{3u}{2}$

in lens displacement method , $u+v=d$ ; $uv=df$

$u+\dfrac{3u}{2}=90$

$u=36$   => $v=54$

$uv=df$ 

$f=\dfrac{36\times 54}{90}=21.6$

option $D$ is correct 

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

In a converging lens of focal length f and the distance between real object and its real image is 4f. If the object moves $x _1$ distance towards lens its image moves $x _2$ distance away from the lens and when object moves $y _1$ distance away from the lens its image moves $y _2$ distance towards the lens, then choose the correct option:-

  1. $x _1>x _2 $ and $y _1>y _2$
  2. $ x _1 < x _2 $ and $ y _1 < y _2 $
  3. $ x _1 < x _2 $ and $y _1>y _2$
  4. $x _1>x _2 $ and $y _2>y _1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a lens with object-image distance 4f, the magnification m = -1 at the center. Near this point, the displacement of the image is greater than the displacement of the object (m > 1 or m < -1), but the question asks about relative movements. Specifically, for a real object and real image, the longitudinal magnification is m^2. Since m^2 > 1 for positions away from 2f, the image moves more than the object.

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

A double convex lens is made of glass which has refractive index $1.55$ for violet rays and $1.50$ for red rays. If the focal length for violet rays is $25$ cm, the focal length for red rays will be nearly

  1. 37.5 cm

  2. 17.5 cm

  3. 27.5 cm

  4. 35 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The lens maker formula states that (1/f) = (n - 1)(1/R1 - 1/R2). For the violet rays, 1/25 = (1.55 - 1) * K, where K represents the curvature term (1/R1 - 1/R2). Solving gives K = 1 / (0.55 * 25). For red rays, 1/f_red = (1.50 - 1) * K = 0.50 * [1 / (0.55 * 25)], which yields f_red = 27.5 cm.

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

The diameter of the sun is $1.4 \times 10 ^ { 9 } \mathrm { m }$ and its distance from the earth is $1.5 \times 10 ^ { 11 } \mathrm { m } .$ The radius of the image of the sun formed by a lens of focal length $20 \mathrm { cm }$ is

  1. $93 mm$
  2. $0.093mm$
  3. $9.3 mm$
  4. $0.93 mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The image size h_i = f * tan(theta), where theta is the angular diameter of the sun. tan(theta) = diameter / distance = 1.4e9 / 1.5e11 = 1.4/150. h_i = 20 cm * (1.4/150) = 200 mm * 0.00933 = 1.86 mm. The radius is half of this, 0.93 mm.

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

The lateral magnification of the lens with an object located at two different positions $u _1$ and $u _2$ are $m _1$ and $m _2$, respectively. Then the focal length of the lens is :

  1. $f=\sqrt {m _1m _2}(u _2-u _1)$
  2. $\dfrac{m _2u _2 - m _1u _1}{m _2-m _1}$
  3. $\dfrac {(u _2-u _1)}{\sqrt {m _2m _1}}$
  4. $\dfrac {(u _2-u _1)}{(m _2)^{-1}-(m _1)^{-1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$u= -u$ ; $f= f$

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$v= \dfrac{fu}{u-f}$

Magnification is: $\dfrac{f}{u-f}$
$\dfrac{m _{1}}{m _{2}}=\dfrac{\frac{f}{u _{1}-f}}{\dfrac{f}{u _{2}-f}}$

$f=\dfrac{u _{2}m _{2}-u _{1}m _{1}}{m _{2}-m _{1}}$

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The ratio of the size of the image to the size of the object is known as :

  1. the focal plane

  2. the transformation ratio

  3. the efficiency

  4. the magnification ratio
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

magnification ratio is given as:

 size or height of image/ size or height of object
substituted with proper sign convention.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A convex lens is used to form an image of an object on a screen. If the upper half of the lens is blackened so that it becomes opaque, then

  1. Only half of the image will be visible

  2. The image position shifts towards the lens

  3. The image position shifts away from the lens

  4. The brightness of the image reduces
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

To form a image only two rays are needed .The total amount of light released by the object is not allowed to pass through the lens, intensity of image will decrease

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The focal length of a convex lens of refractive index $1.5$ is $f$ when it is places in air. When it is immersed in a liquid it behaves as a converging lens its focal length becomes $xf(x>1)$. The refractive index of the liquid

  1. $>3/2$
  2. $<(3/2)$ and $>1$
  3. $<3/2$

  4. all of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac { 1 }{ f } =\left( n-1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right) $
$\Rightarrow \dfrac { 1 }{ f } =\left( \dfrac { 1.5 }{ 1 } -1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right)$ when the lens is placed in air and 
$\dfrac { 1 }{ xf } =\left( \dfrac { 1.5 }{ y } -1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right)$ when the lens is places in the liquid.
where $y=R.l.$ of the liquid
solving we get, $y=\dfrac {3}{2+1/x}$
Hence $(B)$ is correct.

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

Magnification $(m) =$ ______

  1. $\dfrac {v}{u}$
  2. $\dfrac {u}{v}$
  3. $\dfrac {h _{o}}{h _{i}}$
  4. $\dfrac {h _{i}}{h _{o}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Magnification is ratio of the size of the image $h _i$ to the size of the object $h _o$.

$m=\dfrac{h _i}{h _o} = \dfrac{-v}{u}$ where $u$ is object distance and $v$ is image distance.

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

A flim projector magnifies a flim of area $100 $ square centimeter on screen. If linear magnification is $4$ then area of magnified image on screen will be-

  1. $1600 sq. cm$
  2. $800 sq. cm$
  3. $400 sq. cm$
  4. $200 sq. cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As linear magnification, $M=4$

Hence, a real magnification ${ m } _{ r }={ m }^{ 2 }$
${ \left( 4 \right)  }^{ 2 }=16$
Surface area of film image on screen $=16\times 100=1600$ ${ cm }^{ 2 }$.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

Two thin lenses of focal length $f _1$ and $f _2$ are in contact and coaxial. The power of the combination is

  1. $\sqrt{\frac{f _1}{f _2}}$
  2. $\sqrt{\frac{f _2}{f _1}}$
  3. $\frac{f _1+f _2}{f _1f _2}$
  4. $\frac{f _1-f _2}{f _1f _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the lenses placed coaxially, 

We have the relationship $P=P _1+P _2+P _3+......$ 
So,for given case P=$P _1+P _2$ or $P $= $\dfrac{1} {f _1} $+ $\dfrac{1} {f _2} $

So,C is the correct answer.