A dentist obtains a linear magnification of 4 of a hole in a tooth by placing a concave mirror at a distance, of 2.0cm from the tooth. The radius of Curvature of the mirror is

Geraldizuchukwu

7 years ago

Am not convinced with this answer.....
Because since the mirror is concave... The image must be vitual when used as a dentist or shaving mirror and so (F-U) since the image is lesser than the focal length
I chose the option A

Chrisplendour

1 year ago

the answer is 5.3


a concave mirror can produce virtual image when d object is placed beween F and P

Its this position that makes concave mirror suitable for shaving mirror and dentist mirror because when object is placed at that spot, large, virtual,and erect image is formed behind the mirror.

How do i know dat d image is bwn F and P
..i know because bf we can use concave mirror in such application itmust b placed bwn F and C

M=4
U=2
V=?
M=V/U
4=V/2
V=4*2=8
i.e -8 becos its formed bwn F and p

1/F=1/V+1/U
1/F=-1/8 +1/2
1/F=(-1+4)/8
1/F=3/8
F=8/3=2.67

F=r/2
r=2F
r=2*2.67
r=5.3

OsagieN

2 years ago

𝐀 𝐝𝐞𝐧𝐭𝐢𝐬𝐭 𝐮𝐬𝐞𝐬 𝐚 𝐜𝐨𝐧𝐯𝐞𝐫𝐠𝐢𝐧𝐠 𝐦𝐢𝐫𝐫𝐨𝐫 𝐭𝐨 𝐨𝐛𝐭𝐚𝐢𝐧 𝐚𝐧 𝐞𝐫𝐞𝐜𝐭 𝐚𝐧𝐝 𝐦𝐚𝐠𝐧𝐢𝐟𝐢𝐞𝐝 𝐯𝐢𝐫𝐭𝐮𝐚𝐥 𝐢𝐦𝐚𝐠𝐞.

𝐮𝐬𝐢𝐧𝐠 𝐫𝐞𝐚𝐥-𝐢𝐬-𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐜𝐨𝐧𝐯𝐞𝐧𝐭𝐢𝐨𝐧

𝐌 = -𝐟/𝐮-𝐟

𝟰 = -𝐟/𝟮-𝐟

𝟴 -𝟰𝐟 = -𝐟

𝟴 = 𝟯𝐟

𝐟 = 𝟴/𝟯 𝐜𝐦.............

𝐫 = 𝟮𝐟 = 𝟮(𝟴/𝟯) = 𝟭𝟲/𝟯 𝗰𝗺.......𝗔

Choicey09

4 years ago

A dentist obtains a linear magnification of 4 of a hole in a tooth by placing a concave mirror at a distance of 20cm from the tooth. The radius of curvature of the mirror is

Emmanuel24458

2 years ago

To find the radius of curvature \( R \) of the concave mirror, we can use the relationship between magnification \( m \), object distance \( u \), image distance \( v \), and focal length \( f \) of the mirror.

Given:
1. The linear magnification \( m = 4 \)
2. The object distance \( u = -2.0 \) cm (negative because the object is in front of the mirror)
3. The linear magnification formula is \( m = \frac{-v}{u} \), where \( v \) is the image distance.

The magnification \( m \) is also given by \( m = \frac{h'}{h} \), where \( h' \) is the image height and \( h \) is the object height.

Now, we can find \( v \) using the magnification formula:
\[ m = \frac{-v}{u} \]
\[ 4 = \frac{-v}{-2.0} \]
\[ 4 \times (-2.0) = v \]
\[ v = 8.0 \text{ cm} \]

Next, let's use the mirror formula to find the focal length \( f \) of the concave mirror:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
\[ \frac{1}{f} = \frac{1}{-2.0} + \frac{1}{8.0} \]
\[ \frac{1}{f} = -0.5 + 0.125 \]
\[ \frac{1}{f} = -0.375 \]
\[ f = \frac{1}{-0.375} \]
\[ f = -2.67 \text{ cm} \]

The focal length \( f \) is negative, indicating a concave mirror.

The relationship between radius of curvature \( R \) and focal length \( f \) is:
\[ f = \frac{R}{2} \]
\[ R = 2f \]
\[ R = 2 \times (-2.67) \]
\[ R = -5.34 \text{ cm} \]

The radius of curvature \( R \) is positive (since it's the radius of a concave mirror), so:
\[ R = 5.34 \text{ cm} \]

Therefore, the closest answer among the given choices is \( R \approx 5.3 \) cm, so the correct choice is A. 5.3 cm.

Folafoluwa2007

1 year ago

myschool are 100% correct