Numerical Problems of chapter 12: Geometrical Optics

Statements:

1. A concave mirror has a radius of curvature of 15 cm. A small boy of height 90 cm is standing in front of it at 10 cm. Find the location, nature, magnification, and height of the image.

2. A 5 cm tall lighted candle is placed 12 cm from a convex mirror of focal length 6 cm. Find the location, nature, magnification, and height of the image.

3. A girl is standing near a converging mirror of focal length 15 cm. Where should she stand to get:
a. A real, inverted, and three times enlarged image
b. A virtual, erect, and two times diminished image

4. A light ray enters from air to water (n = 4/3) at an angle of incidence of 30°. Find the angle of refraction in water.
If the light ray travels from water to air at an angle of incidence of 40°, find the angle of refraction in air.

5. A light ray is incident from the core (n = 1.42) to the cladding (n = 1.33) at the core-cladding interface. Find the minimum angle at which total internal reflection takes place.

6. A boy is looking at a pen through a concave lens of focal length 10 cm. Find:
a. The power of the lens
b. The location and nature of the image if the pen is at 10 cm from the concave lens

7. Consider a convex lens of focal length 10 cm and a concave lens of focal length 15 cm. Find the power of the lenses.

8. An object is placed in front of a convex lens of power 10 D. From where should the object be placed to get:
a. A three times real image
b. A three times virtual image

Problem 1:

Given:

• Radius of curvature R=15 cm
• Height of the boy h_o=90 cm
• Object distance u=−10 cm (since it’s in front of the mirror)

Solution:

1. Focal length (f) of the concave mirror:

   f=R/2         =15/2              =7.5 cm

2. Mirror Equation:

   1/f=1/v+1/u​

   where $v$ is the image distance and $u$ is the object distance.

   1/7.5=1/v+1/−10​

   Solving for $v$:

   1/v=1/7.5+1/10     =2/15+1/10        =19/30​

   Therefore, v=30/19  ≈1.58 cm

3. Nature of the Image:

   • The positive value of $v$ indicates the image is virtual and formed behind the mirror.

4. Magnification (m):

   m=hi/ho=v/u     m=1.58/−10     ​≈−0.158(negative sign indicates image is virtual and erect)

5. Height of the Image:

   hi=m⋅ho=−0.158×90≈−14.22 cm

Summary:

• Image Location: v≈1.58 cm behind the mirror.
• Nature: Virtual and erect.
• Magnification: −0.158
• Image Height: −14.22 cm(Erect and reduced in size)

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Problem 2:

Given:

• Object height ho=5 cm
• Object distance u=−12 cm
• Focal length f=+6 cm (since it’s a convex mirror)

Solution:

1. Mirror Equation:

   1/f=1/v+1/u      =1/6=1/v+1/−12​

   Solving for $v$:

   1/v=1/6+1/12     =2/12+1/12      =3/12​

   Therefore, v=+4 cm

2. Nature of the Image:

   • The positive value of $v$ indicates the image is virtual and formed behind the mirror.

3. Magnification (m):

   m=hi/ho=v/u       m=4/−12       =−1/3m ​

4. Height of the Image:

   hi=m⋅ho          =−1/3×5       ≈−1.67 cm

Summary:

• Image Location: v=4 cm behind the mirror.
• Nature: Virtual and erect.
• Magnification: −1/3​
• Image Height: −1.67 cm (diminished)

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Problem 3:

Given:

• Focal length of the converging mirror f=15 cm

(a) Real, inverted, and three times enlarged image:

1. Magnification (m) for a real, inverted image is m=−3

2. Using the magnification equation: m=v/u  ⇒ v=−3u

3. Using the mirror equation:

   1/f=1/v+1/u​

   Substituting v=−3u:

   1/15=1/−3u+1/u     ​

   Simplifying:

   1/15=−1+3/3u         =2/3u​

   Therefore:

   u=2/3×15=10 cm

4. Object distance u=10 cm, image distance v=−3u=−30 cm.

Answer:

• Object distance: u=10 cm
• Image distance: v=−30 cm
• The image is real, inverted, and magnified 3 times.

(b) Virtual, erect, and two times diminished image:

1. Magnification (m) for a virtual, erect image is m=+1/2m 

2. Using the magnification equation:

   m=v/u   ⇒   v=1/2u

3. Using the mirror equation:

   1/f=1/v+1/u​

   Substituting v=1/2u:

   1/15=2/u+1/u​

   Simplifying:

   1/15=3/u​

   Therefore:

   u=45 cm

4. Object distance u=45 cm, image distance v=1/2u=22.5 cm.

Answer:

• Object distance: u=45 cm
• Image distance: v=22.5 cm
• The image is virtual, erect, and diminished (half size).

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Problem 4:

Given:

• n_air=1   n_water=4/3​
• Angle of incidence i=30^∘

Solution:

Using Snell’s Law:

n1.sin⁡(i)=n2.sin⁡(r)

1×sin(30∘)=4/3​×sin(r)

sin⁡(r)=1/4/3×sin⁡(30∘)=3/4×1/2=3/8

r=sin^⁡−1(3/8)≈22.02^∘

Answer:

• Angle of refraction r≈22.02^∘

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Light travels from water to air:

• n1=4/3​,              n2=1
• Angle of incidence i=40^∘

Using Snell’s Law:

4/3sin⁡(40^∘)=1×sin⁡(r)

sin⁡(r)=4/3×sin⁡(40^∘)       ≈4/3×0.6428=0.8571

r=sin⁻¹(0.8571)           ≈59.5^∘

Answer:

• Angle of refraction r≈59.5^∘   
  

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Problem 5:

Given:

• Refractive index of the core n1=1.42
• Refractive index of the cladding n2=1.33

Solution: For total internal reflection to occur, the angle of incidence must be greater than the critical angle. The critical angle θc, can be found using the formula:

sin⁡(θc)=n2/n1​​

Substituting the values:

sin⁡(θc)=1.33/1.42   ≈0.9359

θc​=sin⁻¹(0.9359)    ≈68.9^∘

Answer: The minimum angle of incidence for total internal reflection is approximately 68.9°.

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Problem 6:

Given:

• Focal length of the concave lens f=−10 cm (concave lens has a negative focal length)
• Object distance u=−10 cm

Solution:

(a) Power of the lens:

The power P of a lens is given by:

P=1/f(in meters)

Converting focal length to meters:

P=1/−10 cm =1−0.1 m=−10 D

Answer:

• Power of the concave lens: -10 D

(b) Location and Nature of the Image:

Using the mirror equation for the concave lens:

1/f=1/v+1/u​

Substitute f=−10 cm and u=−10 cm:

1/−10=1/v+1/−10​

Simplify:

1/v=1/−10+1/10=0

Thus, the image is formed at infinity (since 1/v=0).

Answer:

• The image is at infinity, virtual, and erect.

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Problem 7:

Given:

• Convex lens focal length f1=10 cm
• Concave lens focal length f2=−15 cm

Solution:

1. Power of the convex lens:

   P1=1/f1=1/10 cm=1/0.1 m=+10 D

2. Power of the concave lens:

   P2=1/f2=1/−15 cm=1/−0.15 m=−6.67 D

3. Total Power:

   P_total=P1+P2             =10 D+(−6.67 D)               =3.33 D

Answer:

• The total power of the lenses is 3.33 D.

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Problem 8:

Given:

• Power of the convex lens P=10 D
• Focal length of the convex lens f=1/P=1/10=10 cm

(a) Object distance for a three times real image:

1. Magnification m=−3(since it’s a real image, the magnification is negative).

2. Using the magnification equation m=v/u, we have:

   v=−3u

3. Using the lens formula:

   1/f=1/v+1/u​

   Substituting v=−3u:

   1/10=1/−3u+1/u​

   Simplifying:

   1/10=−1+3/3u=2/3u​

   Solving for u:

   u=2/3×10=6.67 cm

Answer for (a):

• The object should be placed 6.67 cm from the lens.

(b) Object distance for a three times virtual image:

1. Magnification m=+1/3 (since it’s a virtual image, the magnification is positive).

2. Using the magnification equation m=v/u, we have:

   v=1/3u

3. Using the lens formula:

   1/f=1/v+1/u​

   Substituting v=1/3u:

   1/10=3/u+1/u​

   Simplifying:

   1/10=4/u​

   Solving for $u$:

   u=40 cm

Answer for (b):

• The object should be placed 40 cm from the lens.

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Summary of Solutions:

1. Concave Mirror:

   • Image Location: $1.58\text{cm}$ behind the mirror.
   • Nature: Virtual and erect.
   • Magnification: $-0.158$
   • Image Height: $-14.22\text{cm}$.

2. Convex Mirror:

   • Image Location: $4\text{cm}$ behind the mirror.
   • Nature: Virtual and erect.
   • Magnification: −13-\frac{1}{3}−31​
   • Image Height: $-1.67\text{cm}$.

3. Converging Mirror:

   • (a) Real, inverted, 3 times enlarged image: $u=10\text{cm}$, $v=-30\text{cm}$.
   • (b) Virtual, erect, 2 times diminished image: $u=45\text{cm}$, $v=22.5\text{cm}$.

4. Refraction:

   • Angle of refraction in water: 22.02∘22.02^\circ22.02∘.
   • Angle of refraction from water to air: 59.5∘59.5^\circ59.5∘.

5. Critical Angle: θc=68.9∘\theta_c = 68.9^\circθc​=68.9∘.

6. Concave Lens:

   • Power: $-10\text{D}$.
   • Image: Virtual, erect, formed at infinity.

7. Power of Lenses: $3.33\text{D}$.

8. Object Distance for Lenses:

   • (a) $u=6.67\text{cm}$ for a three times real image.
   • (b) $u=40\text{cm}$ for a three times virtual image.