Question

A 3.21 mm high diamond is placed on the axis of, and 10.5 cm from, a lens with a focal length of -6.79 cm. If it can be determined, is the diamond's image real or virtual? real virtual cannot be determined How high is the image? Express the answer as a positive quantity. image height: mm If it can be determined, is the image upright or inverted with respect to the real thing? inverted upright cannot be determined

Answer 1

Given problem can be solved by using Lens Formula

$1/f=1/v-1/u$

v= negative image Virtual

height of image is can be solved by below formula

magnification=$height of image/height of object$

height is positive image is Upright

Given, High of chiamond, he - † 3.21 mm Distance of diamond u = - 10,5 cm -10,5x10mm = -105 mm -ve means direction of object distance is opposite to light rays direction. tocal length,f=-6.79 cm --6.79x10 mm =-67.9 mm -ve represents the given lens is "Concave lens" distence of diamond Image v= ? Hight of diamond Image, ha= ? From Given deta diamond is placed between focal length (f) and Center of cwivcture (c). Hence diamond Image is formed between focal length (f) and optice Center (0) of concave leur - CS;

distance of diamond Image (V) can se Calculate wring lens Formular Lens formula > 0. substitute t= -67.9 mm . u= -105 mm in equation -67-9 = +- Gtos) -67-9 = + + tos *=-6719-vos =-602.9+ atos) - the thing) = -6730) - E3 11 105 +67.9 67.9x105 1130 - Hlmm -17130 173 : distance of diamond Image, V = -41 mm/ Here diamond Image is virtual - since was are sign Image height: magnification, ma hoi - ñ → © susstilite ho=+3.21 mm V=-Himm u= - 105 mm in equition ②

105 h; - - * 312) 13146-1.25 mm Image height, hi =+1125 mm Here he value is Positive" hence Image of diamond is /upright] mau 105 Image is dininished Imic1 heure