HEY BRAINLIANS AND MODERATORS❤SOLVE THIS QUESTION❤{IF POSSIBLE SOL

1.	HEY BRAINLIANS AND MODERATORS❤SOLVE THIS QUESTION❤{IF POSSIBLE SOLVE IT IN PAPER}Include the respective diagram to it and No shortcuts solve it properly ⚠️⛔SPAMMER STAY AWAY FROM THE QUESTION FOR ABOUT 1000 LAKH LIGHT YEARS⛔⚠️THANK YOU
Answer» QUESTION: A telescope has an OBJECTIVE LENS of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a DISTANCE of 1 km is observed through this telescope in normal SETTING, the angle formed by the image of the tower is θ , then θ is close to ANSWER: θ is close to 60°. GIVEN: Focal length of telescope(F) = 150 cm Focal length of eyepiece(f) = 50 cm Distance from the tower(d) = 1000m Height of the tower = 50m TO FIND: CLOSEST VALUE OF θ FORMULA: Magnification from mirror: $Magnifying\:\:distance(m)= F/f$ EXPLANATION: d = 1000 m F = 150 cm f = 50 cm $Magnifying\:\:distance =150 / 5= 30\\ \\ \\ Magnifying\:\:distance = \dfrac{\tan\beta}{\tan\alpha } = 30\\ \\ \\\tan \alpha = \dfrac{f}{d}\\ \\ \\ \tan \alpha = \dfrac{50}{1000} \\ \\ \\ \tan \alpha = \dfrac{1}{20} \\ \\ \\ \tan \beta = m\times \tan \alpha \\ \\ \\ \tan \beta = 30\times \dfrac{1}{20} \\ \\ \\ \tan\beta = \dfrac{30}{20} \\ \\ \\ \tan \beta = \dfrac{3}{2}\:\:radians \\ \\ \\ \beta =tan^{-1}\bigg( \dfrac{3}{2}\bigg) \\ \\ \\ \beta \approx 60^{ \circ}$ $\bold{\large{NOTE 1 : \tan \alpha= Angle\:\:formed\:\:by\:\:the \:\:eyepiece}}$ $\bold{\large{NOTE 2 :\tan \beta =Angle\:\:formed\:\:by\:\:the \:\:image}}$ OTHER FORMULA: $m= H/h= -v/ u\\ \\ \\H= Height\:\:of\:\: image\\ \\\\h= Height\:\:of\:\:object\\ \\ \\u= Object\:\:distance\\ \\ \\v= Image\:\:distance$

HEY BRAINLIANS AND MODERATORS❤SOLVE THIS QUESTION❤{IF POSSIBLE SOLVE IT IN PAPER}Include the respective diagram to it and No shortcuts solve it properly ⚠️⛔SPAMMER STAY AWAY FROM THE QUESTION FOR ABOUT 1000 LAKH LIGHT YEARS⛔⚠️THANK YOU

Answer»

QUESTION:

A telescope has an OBJECTIVE LENS of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a DISTANCE of 1 km is observed through this telescope in normal SETTING, the angle formed by the image of the tower is θ , then θ is close to

ANSWER:

θ is close to 60°.

GIVEN:

Focal length of telescope(F) = 150 cm

Focal length of eyepiece(f) = 50 cm

Distance from the tower(d) = 1000m

Height of the tower = 50m

TO FIND:

CLOSEST VALUE OF θ

FORMULA:

Magnification from mirror:

$Magnifying\:\:distance(m)= F/f$

EXPLANATION:

d = 1000 m

F = 150 cm

f = 50 cm

$Magnifying\:\:distance =150 / 5= 30\\ \\ \\ Magnifying\:\:distance = \dfrac{\tan\beta}{\tan\alpha } = 30\\ \\ \\\tan \alpha = \dfrac{f}{d}\\ \\ \\ \tan \alpha = \dfrac{50}{1000} \\ \\ \\ \tan \alpha = \dfrac{1}{20} \\ \\ \\ \tan \beta = m\times \tan \alpha \\ \\ \\ \tan \beta = 30\times \dfrac{1}{20} \\ \\ \\ \tan\beta = \dfrac{30}{20} \\ \\ \\ \tan \beta = \dfrac{3}{2}\:\:radians \\ \\ \\ \beta =tan^{-1}\bigg( \dfrac{3}{2}\bigg) \\ \\ \\ \beta \approx 60^{ \circ}$

$\bold{\large{NOTE 1 : \tan \alpha= Angle\:\:formed\:\:by\:\:the \:\:eyepiece}}$

$\bold{\large{NOTE 2 :\tan \beta =Angle\:\:formed\:\:by\:\:the \:\:image}}$

OTHER FORMULA:

$m= H/h= -v/ u\\ \\ \\H= Height\:\:of\:\: image\\ \\\\h= Height\:\:of\:\:object\\ \\ \\u= Object\:\:distance\\ \\ \\v= Image\:\:distance$