Numerical Problems 12.1: An object 10.0 cm in front of a convex mirror forms an image 5.0 cm behind the mirror.  What is the focal length of the mirror? Ans. (10 cm)

... 


Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions.
For easy and comprehensive understanding of sign convention of spherical mirrors problems please Click on the below link


Given Data:


Distance of the object from the mirror = p = -10 cm

Distance of the image from the mirror = q  = 5 cm 

Note: You can use notations  u and v instead of p and q depending upon the notation used in your book.


To Find:


Focal Length = f  = ?   



Solution:

We have the Spherical Mirror formula as 

`\frac{1}{f}` = `\frac{1}{p}` + `\frac{1}{q}`

 Now by putting the values

`\frac{1}{f}` = `\frac{1}{-10cm}` + `\frac{1}{5cm}`

`\frac{1}{f}` = -`\frac{1}{10cm}` + `\frac{1}{5cm}`

`\frac{1}{f}` = `\frac{-1 + 2}{10cm}` 

`\frac{1}{f}` = `\frac{1}{10cm}`

by cross multiplication we get

f = 10 cm ----------------Ans.

(for convex mirror focal length is always Positive) 



Numerical Problems 12.2: An object 30 cm tall is located 10.5 cm from a concave mirror with focal length 16 cm. (a) Where is the image located? (b) How high is it? Ans. [ (a) 30.54 cm (b) 87.26 cm]

Ans. (1.63 msᐨ²)

... 


Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions.
For easy and comprehensive understanding of sign convention of spherical mirrors problems please Click on the below link


Given Data:


Height of the object: h₀ = 30 cm

Distance of the object from the mirror = p = -10 cm

Focal Length = f  = -16 cm


Note: You can use notations u and v instead of p and q depending upon the notation used in your book.


To Find:


(a) Distance of the image from the mirror = q  = ?

(b) Image height = hᵢ = ? 



Solution:


(a) Distance of the image from the mirror = q  = ?


We have the Spherical Mirror Formula as 

`\frac{1}{f}` = `\frac{1}{p}` + `\frac{1}{q}`

or

`\frac{1}{q}` = `\frac{1}{f}` - `\frac{1}{p}`

 Now by putting the values

`\frac{1}{q}` = `\frac{1}{-16cm}` - `\frac{1}{-10.5cm}`

`\frac{1}{q}` = -`\frac{1}{16cm}` + `\frac{10}{105cm}`

`\frac{1}{q}` = `\frac{-105+160}{(16)(105) cm}`

`\frac{1}{q}` = `\frac{55}{(16)(105) cm}`

by simplifying and flipping the fraction we get

q =  30.54 cm ----------------Ans. 1

the positive sign shows that image is virtual and formed behind the concave mirror.




(b) Image height = hᵢ = ? 

We have the magnification formula as 

 magnification = `\frac{hᵢ }{h₀}` = `\frac{q}{p}`

or

`\frac{hᵢ }{h₀}` = `\frac{q}{p}`

by putting values

`\frac{hᵢ}{30 cm}` =`\frac{30.54 cm}{-10.5 cm}`

hᵢ = -`\frac{30.54 cm}{10.5 cm}` 𝐱 30 cm

By Simplifying we get

hᵢ = -87.26 cm---------------Ans. 2

the negative sign shows that the image is inverted



Numerical Problems 12.3: An object and its image in a concave mirror are of the same height, yet inverted, when the object is 20 cm from the mirror. What is the focal length of the mirror? Ans. (10 cm)

... 


Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions.
For easy and comprehensive understanding of sign convention of spherical mirrors problems please Click on the below link



Given Data:


The given condition can be illustrated in the below image, So



Distance of the object from the mirror = p = -20 cm

Distance of the image from the mirror = q = -20 cm


Note: You can use notations u and v instead of p and q depending upon the notation used in your book.



To Find:


Focal Length = f  = ?   


Solution:


We have the mirror formula as 

`\frac{1}{f}` = `\frac{1}{p}` + `\frac{1}{q}`

 Now by putting the values

`\frac{1}{f}` = `\frac{1}{-20cm}` + `\frac{1}{-20cm}`

`\frac{1}{f}` = -`\frac{1}{20cm}` - `\frac{1}{20 cm}`

`\frac{1}{f}` = `\frac{-1 - 1}{20cm}` 

`\frac{1}{f}` = `\frac{-2}{20cm}`

`\frac{1}{f}` = -`\frac{1}{10cm}`

by cross multiplication we get

f = -10 cm  ----------------Ans.

The negative sign just shows that the focal length distance is measured on front side of concave mirror


Numerical Problems 12.4: Find the focal length of a mirror that forms an image 5.66 cm behind the mirror of an object placed at 34.4 cm in front of the mirror. Is the mirror concave or convex? Ans. ( 6.77 cm, Convex mirror)

... 

Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions.
For easy and comprehensive understanding of sign convention of spherical mirrors problems please Click on the below link


Given Data:

Distance of the object from the mirror = p = - 34.4 cm 
Distance of the image from the mirror = q  =  5.66 cm

Note: You can use notations u and v instead of p and q depending upon the notation used in your book. 


To Find:

Focal Length = f  =  ?  
 

Solution:

We have the mirror formula as 

`\frac{1}{f}` = `\frac{1}{p}` + `\frac{1}{q}`

Now by putting the values

`\frac{1}{f}` = `\frac{1}{-34.4 cm}` + `\frac{1}{5.66 cm}`

`\frac{1}{f}` = -`\frac{1}{34.4}` cm + `\frac{1}{5.66}` cm

`\frac{1}{f}` = -`\frac{10}{344}` cm + `\frac{100}{566}` cm

`\frac{1}{f}` = `\frac{-(10)(566) + (100)(344)}{(344)(566) }` cm

`\frac{1}{f}` = `\frac{-5,660 + 34,400}{194,704}` cm

`\frac{1}{f}` = `\frac{ 28,740}{194,704}` cm

or

`\frac{f}{1}` = `\frac{194,704}{28,740}` cm

by simplifying we get

f =  6.77 cm  -----------------------Ans.

The Positive sign of focal length shows that it is a Convex Mirror




Numerical Problems 12.5: An image of a statue appears to be 11.5 cm behind a convex mirror with focal length 13.5 cm. Find the distance from the statue to the mirror. Ans. (77.62 cm)

... 

Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions.
For easy and comprehensive understanding of sign convention of spherical mirrors problems please Click on the below link


Given Data:

Distance of the image from the mirror = q  = 11.5 cm 

Focal Length = f = 13.5 cm 

Note: You can use notation u and v instead  of p  and q depending  upon the notation used in your book.


To Find:

Distance of the object from the mirror = p = ? 

Solution:

We have the mirror formula as 

`\frac{1}{f}` = `\frac{1}{p}` + `\frac{1}{q}`

or

`\frac{1}{p}` = `\frac{1}{f}` - `\frac{1}{q}`

 Now by putting the values

`\frac{1}{p}` = `\frac{1}{13.5 cm}` - `\frac{1}{11.5 cm}`

`\frac{1}{p}` = `\frac{10}{135}` cm - `\frac{10}{115cm}` cm

`\frac{1}{p}` = `\frac{(10)(115) - (10)(135)}{(135)(115) }` cm

`\frac{1}{p}` = `\frac{1,150 - 1,350}{15,525}` cm

`\frac{1}{p}` = `\frac{-200}{15,525}` cm

or

`\frac{p}{1}` = -`\frac{15,525}{20}` cm

by simplifying we get

p = - 77.625 cm ------------------Ans.

the -ve sign shows that the object of on left side of the mirror. 

Numerical Problems 12.6: An image is produced by a concave mirror of focal length 8.7 cm. The object is 13.2 cm tall and at a distance 19.3 cm from the mirror. 

(a) Find the location and height of the image. 
(b) Find the height of the image produced by the mirror if the object is twice as far from the mirror. 

Ans. [(a) 15.84 cm, 10.83 cm (b) 5.42 cm] 

... 

Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions.
For easy and comprehensive understanding of sign convention of spherical mirrors problems please Click on the below link


Given Data:


Focal Length = f  = - 8.70 cm

Height of of the object: h₀ = 30 cm

Distance of the object from the mirror = p = -19.3 cm

Note: You can use notations u and v instead of p and q depending upon the notation used in your book.



To Find:


(a) 
Distance of the image from the mirror = q  = ?
Image height = hᵢ = ? 

(b) 
Image height = hᵢ = ? (when the object is twice far from the mirror)


Solution:


(a)
We have the mirror formula as 

`\frac{1}{f}` = `\frac{1}{p}` + `\frac{1}{q}`

or

`\frac{1}{q}` = `\frac{1}{f}` - `\frac{1}{p}`

Now by putting the values

`\frac{1}{q}` = `\frac{1}{-8.7 cm}` - `\frac{1}{-19.3 cm}`

`\frac{1}{q}` = -`\frac{10}{87}` cm + `\frac{10}{193}` cm

`\frac{1}{q}` = `\frac{-1930+870}{(87)(193)}` cm

`\frac{1}{q}` = `\frac{-1,060}{16,791}` cm

by reversing the fraction we get

`\frac{q}{1}` = -`\frac{16,791}{1,060}` cm

by simplifying we get

q = - 15.83 cm ---------------Ans. 1(a)



Now using the magnification formula

magnification = `\frac{hᵢ }{h₀}` = `\frac{q}{p}`

or

`\frac{hᵢ }{h₀}` = `\frac{q}{p}`

by putting values

`\frac{hᵢ}{13.2 cm}` =`\frac{-15.84 cm}{-19.3 cm}`

hᵢ = `\frac{15.84 cm}{19.3 cm}` 𝐱 13.2 cm

by simplifying we get

hᵢ = 10.83 cm ---------------Ans. 2(a)



(b)
when the object is twice far from the mirror i.e
P = -19.3 cm  𝐱 2 = - 38.6 cm

Using the magnification formula

`\frac{hᵢ }{h₀}` = `\frac{q}{p}`

by putting values

`\frac{hᵢ}{13.2 cm}` =`\frac{-15.84 cm}{-38.6 cm}`

hᵢ = `\frac{15.84 cm}{38.6 cm}` 𝐱 13.2 cm

by simplifying we get

hᵢ = 5.42 cm



Numerical Problems 12.7: Nabeela uses a concave mirror when doing her makeup. The mirror has a radius of 12.11 curvature of 38 cm. 

(a) What is the focal length of the mirror? 
(b) Nabeela is located 50 cm from the mirror. Where will her image appear?
(c) Will the image be upright or inverted? 
 Ans. [(a) 19 cm, (b) 30.64 cm, (c) Inverted]

... 

Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions.
For easy and comprehensive understanding of sign convention of spherical mirrors problems please Click on the below link


Given Data:


Nature of Mirror used = Concave (Converging) 

Radius of Curvature = R= - 38 cm 

Distance of the object from the mirror = p = - 50 cm


[ Note: You can use u and v instead of p and q depending upon the notation used in your book. ]


To Find:


(a) Focal length = f = ?

(b) Distance of the image from the mirror = q  = ?

(c) Nature of the Image height =  ? 




Solution:


(a) Focal length = f = ?

We have the formula for focal length when radius of curvature are known as,

Focal Length = f = `\frac{-R}{2}`= 

Focal Length = f =-`\frac{38}{2}` 

Focal Length = f = -19 cm -----------Ans. 1




(b) Distance of the image from the mirror = q  = ?


We have the mirror formula as 

`\frac{1}{f}` = `\frac{1}{p}` + `\frac{1}{q}`

or

`\frac{1}{q}` = `\frac{1}{f}` - `\frac{1}{p}`


Now by putting the values


`\frac{1}{q}` = `\frac{1}{-19 cm}` - `\frac{1}{-50 cm}`

`\frac{1}{q}` = -`\frac{1}{-19 cm}` + `\frac{1}{-50 cm}`

`\frac{1}{q}` = `\frac{-50+19}{(19)(50)}` cm

`\frac{1}{q}` = `\frac{-31}{1,950}` cm

by flipping the fraction both side we get

`\frac{q}{1}` = -`\frac{1,950}{31}` cm

by simplifying we get

q = - 30.64 cm


(c) Inverted:

According to the above results, the object is placed behind the center of curvature, an image is formed between the center of curvature and focus. The size of the image is smaller than compared to that of the object but the image is real and inverted. This case can be illustrated as below



Numerical Problems 12.8: An object 4 cm high is placed at a distance of 12 cm from a convex lens of focal length 8 cm. Calculate the position and size of the image. Also state the nature of the image. Ans. (24 cm, 8 cm, image is real, inverted and magnified)

... 

Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions. 
For easy and comprehensive understanding of sign convention of spherical lenses in Numerical problems please Click on the below link 
[New Cartesian Sign Convention for Spherical Lenses}


Given Data:

Nature of lens used = Convex (Converging)

Height of of the object: h₀ = 4 cm

Distance of the object from the mirror = p = -12 cm

Focal Length = f  = 8 cm

Note: You can use notations u and v instead of p and q depending upon the notation used in your book.



To Find:

Distance (position) of the image from the mirror = q  = ?

Image height (size) = hᵢ = ? 

Nature of the Image = ? 



Solution:


We have the lens formula according to the New Cartesian Sign Convention as 

`\frac{1}{f}` = `\frac{1}{q}` - `\frac{1}{p}`

or

`\frac{1}{q}` = `\frac{1}{f}` + `\frac{1}{p}`

Now by putting the values

`\frac{1}{q}` = `\frac{1}{8 cm}` + `\frac{1}{-12 cm}`

`\frac{1}{q}` = `\frac{1}{8 cm}` - `\frac{1}{12 cm}`

`\frac{1}{q}` = `\frac{12-8}{(8)(12)}` cm

`\frac{1}{q}` = `\frac{4}{96}` cm

by reversing the fraction we get

`\frac{q}{1}` = `\frac{96}{4}` cm

by simplifying we get

q = 24 cm  -----------------Ans.1

The Positive distance shows that the image is real.




Now using the magnification formula

magnification = `\frac{hᵢ }{h₀}` = `\frac{q}{p}`

or

`\frac{hᵢ }{h₀}` = `\frac{q}{p}`

by putting values

`\frac{hᵢ}{4 cm}` =`\frac{24 cm}{-12 cm}`

hᵢ = -`\frac{24 cm}{12 cm}` 𝐱 4 cm

by simplifying we get

hᵢ = - 8 cm  --------------Ans. 2

The negative sign shows that the image is inverted

Ans. 3 : As from the above results i.e. hᵢ > h₀ and the lens is convex, so the image formed is Real, Inverted and magnified. 


Numerical Problems 12.9: In object 10 cm high is placed at a distance of 20 cm from a concave lens of focal length 15 cm. Calculate the position and size of the image. Also, state the nature of the image. Ans. (-8.57 cm, 4.28 cm, image is virtual, erect and diminished

... 

Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions. 
For easy and comprehensive understanding of sign convention of spherical lenses in Numerical problems please Click on the below link 
[New Cartesian Sign Convention for Spherical Lenses}


Given Data:

Nature of lens used = Concave (Diverging)

Height of of the object: h₀ = 10 cm

Distance of the object from the lens = p = -20 cm

Focal Length = f  = -15 cm


[ Note: You can use u and v instead of p and q depending upon the notation used in your book. ]


To Find:


Distance (position) of the image from the Lens = q  = ?

Image height (size) = hᵢ = ? 

Nature of the Image = ? 



Solution:

We have the General lens formula as according new Cartesian sign convention.  

`\frac{1}{f}` = `\frac{1}{q}` - `\frac{1}{p}`

or

`\frac{1}{q}` = `\frac{1}{p}` + `\frac{1}{f}`


Now by putting the values

`\frac{1}{q}` = `\frac{1}{-20 cm}` + `\frac{1}{-15 cm}`

`\frac{1}{q}` = -`\frac{1}{15 cm}` - `\frac{1}{20 cm}`

`\frac{1}{q}` = `\frac{-20-15}{(15)(20)}` cm

`\frac{1}{q}` = `\frac{-35}{300}` cm

by reversing the fraction we get

`\frac{q}{1}` = -`\frac{300}{35}` cm

by flipping and simplifying we get

q = -8.57 cm -------------Ans. 1

the negative sign shows that an image is virtual




Now using the magnification formula

magnification = `\frac{hᵢ }{h₀}` = `\frac{q}{p}`

or

`\frac{hᵢ }{h₀}` = `\frac{q}{p}`

by putting values

`\frac{hᵢ}{-10 cm}` =`\frac{-8.57 cm}{-20 cm}`

hᵢ = `\frac{-8.57 cm}{-20 cm}` 𝐱 (-10 cm)

by simplifying we get

hᵢ = - 4.28 cm -------------Ans. 2

The negative shows that image is inverted 

Ans. 3. 
hᵢ < h₀ and the Concave lens shows that image formed is Virtual and diminished.



Numerical Problems 12.10: A convex lens of focal length 6 cm is to be used to form a virtual image three times the size of the object. Where must the lens be placed? Ans. (-4 cm)

... 

Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions. 
For easy and comprehensive understanding of sign convention of spherical lenses in Numerical problems please Click on the below link 
[New Cartesian Sign Convention for Spherical Lenses}


Given Data:

Nature Of lens used = Convex (Converging) 
Focal Length = f = 6 cm 
Let the object distance = -p cm then
Distance of the image from the lens = q  = - 3 (-p) = 3p (The first -ve sign is being of a virtual image) 
 

To Find:

Distance of the object from the lens = p = (-ve) ? 

Solution:

We have the General lens formula as according new Cartesian sign convention.  

`\frac{1}{f}` = `\frac{1}{q}` - `\frac{1}{p}`

by putting values

`\frac{1}{6}` cm = `\frac{1}{3p}` - `\frac{1}{p}`

`\frac{1}{6}` cm = `\frac{1}{3p}` - `\frac{1}{p}`

`\frac{1}{6}` cm = `\frac{1 -  3}{3p}`

`\frac{1}{6}` cm = `\frac{-2}{3p}`

or

-`\frac{2}{3p}`= `\frac{1}{6}` cm

Now reversing (Flipping) the fraction and also multiplying by -1 both side we get,

`\frac{3p}{2}` =  - `\frac{6}{1}` cm

or 

p = - `\frac{6}{1}` x `\frac{2}{3}` cm

by simplifying we get

p = - 4 cm --------------Ans.

The negative shows that object is real and all real object distance are taken as negative according to new Cartesian conversion rules.



Numerical Problems 12.11: A ray of light from air is incident on a liquid surface at an angle of incidence 35⁰. Calculate the angle of refraction if the refractive index of the liquid is 1.25. Also calculate the critical angle between the liquid air inter-face. Ans. (27.31⁰, 53.13⁰)

...

Given Data:


Angle of incidence = i = 35⁰

Refractive Index = n  = 1.25



To Find:


Angle of refraction = r = ?

Critical Angle = C = ?




Solution:


According to the Snell's Law

`\frac{sin i}{sin r}` = n

or

sin r = `\frac{sin i}{n}`

by putting values

sin r = `\frac{sin (35⁰)}{1.25}`

sin r = `\frac{0.57}{1.25}`

sin r = 0.456
or
r = sin⁻¹ (0.456)

r = 27.13⁰ ---------------Ans. 1




The formula for critical angle (C) as following

sin C = `[\frac{1}{n}]`

or 

C = sin⁻¹`[\frac{1}{n}]`

C = sin⁻¹`[\frac{1}{1.25}]`

C = sin⁻¹(0.8)

C = 52.13⁰ ---------------Ans.2



Numerical Problems 12.12:The power of a convex lens is 5 D. At what distance the object should be placed from the lens so that its real and 2 times larger image is formed. Ans. (-30 cm)

...

Note: In this numerical all the sign conventions are according to New Cartesian Sign Conventions. 
For easy and comprehensive understanding of sign convention of spherical lenses in Numerical problems please Click on the below link 
[New Cartesian Sign Convention for Spherical Lenses}


Given:

Nature of Lens = Convex (Converging)
Power of lens = P = 5D 
Let the object distance = p cm
Distance of the image from the lens = q  = 2p (+ve because of real image) 

To Find:

Distance of the object from the lens = p = ? (-ve)

Solution: 

We have the General lens formula as according new Cartesian sign convention.  

`\frac{1}{f}` = `\frac{1}{q}` - `\frac{1}{p}`

But the values of focal length (f) is also unknown, so to find focal length (f) we know that 

Power of lens = P = `\frac{1}{f}`

or 

f = `\frac{1}{P}`

f = `\frac{1}{5}`

f = 0.2 m

or 

f = 0.2 𝐱 100 cm

f = 20 cm (+ve for Convex lense)

We have the lens formula as

`\frac{1}{f}` = `\frac{1}{q}` - `\frac{1}{p}`

by putting the values

`\frac{1}{20 cm}` = `\frac{1}{2p}` - `\frac{1}{p}`

`\frac{1}{20}` cm = `\frac{-2 - 1}{2p}`

`\frac{1}{20}` cm = `\frac{-3}{2p}`

`\frac{1}{20}` cm = -`\frac{3}{2p}`

or

-`\frac{3}{2p}`= `\frac{1}{20}` cm

or

-`\frac{2p}{3}` = `\frac{20}{1}` cm

p = `\frac{20}{1}` x `\frac{-3}{2}` cm

by simplifying we get

p = -30 cm

The object distance is -30 cm on the left side of the lens and is taken -ve because of the new Cartesian conventions.