Question

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QUESTION :---

(1.) A convex Mirror Used For Rear - View on an automobile has a radius of Curvature of 3.00 m. If a bus is Located at 5.00 m from this mirror. Find the Position, nature and size of the image.

(2.) Rules of Refraction with diagram (In copy).

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Answer 1

1.) QUESTION : ᴀ ᴄᴏɴᴠᴇx ᴍɪʀʀᴏʀ ᴜsᴇᴅ ғᴏʀ ʀᴇᴀʀ - ᴠɪᴇᴡ ᴏɴ ᴀɴ ᴀᴜᴛᴏᴍᴏʙɪʟᴇ ʜᴀs ᴀ ʀᴀᴅɪᴜs ᴏғ ᴄᴜʀᴠᴀᴛᴜʀᴇ ᴏғ 3.00 ᴍ. ɪғ ᴀ ʙᴜs ɪs ʟᴏᴄᴀᴛᴇᴅ ᴀᴛ 5.00 ᴍ ғʀᴏᴍ ᴛʜɪs ᴍɪʀʀᴏʀ. ғɪɴᴅ ᴛʜᴇ ᴘᴏsɪᴛɪᴏɴ, ɴᴀᴛᴜʀᴇ ᴀɴᴅ sɪᴢᴇ ᴏғ ᴛʜᴇ ɪᴍᴀɢᴇ.

GIVEN :

• Radius of the Curvature = 3.00m
• Location of the bus from the mirror = 5.00m

TO FIND :

• The Position, Nature and the size of the image = ?

STEP - BY - STEP EXPLAINATION :

➠ Radius of the Curvature (R) = 3.00 m

➠ Distance of the Object { mirror } (u) = -5.00m

➠ Distance of the Image (v) = ?

➠ Height of the Image (h') = ?

➠ Forcal Length (f) = Radius / 2 = + 3.00 m / 2 = + 1.50

➠ 1/v + 1/u = 1/f (Assuming per as Formula)

$➠\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = + \frac{1}{1.50} - \frac{1}{( - 5.00)} = \frac{1}{1.50} + \frac{1}{5.00} = \frac{5.00 + 1.50}{7.50}$

$➠ \: v = \frac{ + 7.50}{6.50} = + 1.15$

Therefore, The image is 1.15 m at the back of the mirror.

$➠ \: m = \frac{h'}{h} = \frac{v}{u} = - \frac{1.15}{ - 5.00} = + 0.23$

The image will be formed as :---

• Errect
• Virtual
• Smaller in Size

with a factor of 0.23

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2.) QUESTION : ʀᴜʟᴇs ᴏғ ʀᴇғʀᴀᴄᴛɪᴏɴ ᴡɪᴛʜ ᴅɪᴀɢʀᴀᴍ ɪɴ ᴛʜᴇ ᴄᴏᴘʏ.

ANSWER :

(1.) A ray of light bends towards the 'Normal' when when it goes (obliquely) from an optical rarer medium to an optical density medium.

(2.) A Ray of light bends away from the 'Normal' when it goes (obliquely) from an optically denser medium to an optically rarer medium.

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