CBSE Class 12-science Answered

An angular magnification of30X is desired using objectve of focal length1.25 cm and an eye peice of focal length 5 cm. How would you set up compound icroscope

Asked by yashjha23 | 08 Jan, 2016, 01:12: AM

Expert Answer

In normal adjustment of microscope, the image is formed atthe least distance of distinct vision D = 25 cm.

Angular magnification of the eye piece,

begin mathsize 14px style straight m subscript straight e equals 1 plus straight D over straight f equals 1 plus 25 over 5 equals 6 end style

Given total angular magnification, M (=m₀×m_e)= 30

begin mathsize 14px style therefore end style

Magnification of objective,

begin mathsize 14px style straight m subscript 0 equals straight M over straight m subscript straight e equals 30 over 6 equals 5 end style

The objective lens of microscope forms real image, so its magnification is negative.

begin mathsize 14px style straight v subscript straight o over straight u subscript straight o equals negative 5 straight v subscript straight o equals negative 5 space straight u subscript straight o end style

As u_o is negative and v_o is positive.

begin mathsize 14px style straight u subscript straight o equals negative open vertical bar straight u subscript straight o close vertical bar straight v subscript straight o equals plus 5 open vertical bar straight u subscript straight o close vertical bar end style

Focal length of the objective lens, f₀ = 1.25 cm

From the lens formula,

begin mathsize 14px style 1 over straight f subscript straight o equals 1 over straight v subscript straight o minus 1 over straight u subscript straight o fraction numerator 1 over denominator 1.25 end fraction equals fraction numerator 1 over denominator 5 open vertical bar straight u subscript straight o close vertical bar end fraction plus fraction numerator 1 over denominator open vertical bar straight u subscript straight o close vertical bar end fraction open vertical bar straight u subscript straight o close vertical bar equals 6 over 5 cross times 1.25 space cm equals 1.5 space cm Also comma space open vertical bar straight v subscript straight o close vertical bar equals 5 open vertical bar straight u subscript straight o close vertical bar equals 5 cross times 1.5 equals 7.5 space cm end style

Focal length of the eye, f_e=5 cm

From lens formula,

begin mathsize 14px style 1 over straight f subscript straight e equals 1 over straight v subscript straight e minus 1 over straight u subscript straight e 1 over straight u subscript straight e equals 1 over straight v subscript straight e minus 1 over straight f subscript straight e equals negative 1 over 25 minus 1 fifth equals negative 6 over 25 straight u subscript straight e equals negative 25 over 6 equals negative 4.17 space cm open vertical bar straight u subscript straight e close vertical bar equals 4.17 space cm Separation space between space the space lenses comma space straight d equals open vertical bar straight v subscript straight o close vertical bar plus open vertical bar straight u subscript straight e close vertical bar equals 7.5 plus 4.17 equals 11.67 space cm end style

Thus to obtain the desired magnification, the separation between the lenses must be 11.67 cm and the objective lens must be placed at a distance of 1.5 cm in front of the objective lens.

Answered by Faiza Lambe | 08 Jan, 2016, 09:27: AM